OI常用模板

 1 long long qpow(long long a,long long b,int mod)
 2 {
 3     long long res=1;
 4     while (b)
 5     {
 6         if (b&1) res=res*a%mod;
 7         a=a*a%mod;
 8         b>>=1;
 9     }
10     return res%mod;
11 }

快速幂

 1 const int mod1=19260817;
 2 const int mod2=23333333;
 3 const int mod3=33333331;
 4 bool show[mod3];
 5 int hash(string a)
 6 {
 7     long long val=1;
 8     int ha1=0,ha2=0;
 9     int len=a.length();
10     for(register int i=len-1;i>-1;i--)
11     {
12         ha1+=a[i]*val;
13         ha1%=mod1;
14         ha2+=a[i]*val;
15         ha2%=mod2;
16         val*=26;
17         if(val > 10000000 ) val=331;
18     }
19     return (long long)ha1*ha2%mod3;
20 }

字符串哈希

 1 int k,n,m,cnt,sum,ai,bi,ci,head[5005],dis[5005],vis[5005];
 2 struct Edge
 3 {
 4     int v,w,next;
 5 }e[400005];
 6 void add(int u,int v,int w)
 7 {
 8     e[++k].v=v;
 9     e[k].w=w;
10     e[k].next=head[u];
11     head[u]=k;
12 }
13 typedef pair <int,int> pii;
14 priority_queue <pii,vector<pii>,greater<pii> > q;
15 int Prim()
16 {
17     memset(dis,127,sizeof(dis));
18     memset(head,-1,sizeof(head));
19     dis[1]=0;
20     q.push(make_pair(0,1));
21     while(!q.empty()&&cnt<n)
22     {
23         int d=q.top().first,u=q.top().second;
24         q.pop();
25         if(vis[u]) continue;
26         cnt++;
27         sum+=d;
28         vis[u]=1;
29         for(R i=head[u];i!=-1;i=e[i].next)
30             if(e[i].w<dis[e[i].v])
31                 dis[e[i].v]=e[i].w,q.push(make_pair(dis[e[i].v],e[i].v));
32     }
33     if (cnt==n) return sum;
34     return -1;

Prim

 1 struct Edge
 2 {
 3     int x,y,z;
 4 }e[200001];
 5 int n,m,u,v,fu,fv,cnt,ans;
 6 int fa[200001];
 7 bool cmp(Edge a,Edge b){return a.z < b.z || (a.z==b.z && a.x<b.x);}
 8 int find(int x){return x == fa[x]?x:fa[x]=find(fa[x]);}
 9 void kruskal()
10 {
11     for(int i=1;i<=m;i++)  fa[i] = i;
12     for(int i=1;i<=n;i++)
13     {
14         u = e[i].x;
15         v = e[i].y;
16         fu = find(u);
17         fv = find(v);
18         if(fu != fv)
19         {
20             fa[fu] = fv;
21             ans += e[i].z;
22             cnt++;
23         }
24         if(cnt == m-1) break;
25     }
26     if (ans) return ans;
27     return -1'       
28 }

Kruskal(最后return,";"写成"'"了，请自行改过来)

 1 #ifndef DS
 2 #define DS
 3 template <unsigned const long long maxn=20>
 4 class i_Ds
 5 {
 6     typedef long long X;
 7     X s[maxn],height[maxn];
 8     typedef X inte;
 9     public:
10         void init(){for (inte i=0;i<maxn;i++) s[i]=i,height[i]=0;}
11         i_Ds(){init();}
12         X find(X x) {if (x!=s[x]) s[x]=find(s[x]);return s[x];}
13         void _union(X x,X y)
14         {
15             x=find(x);y=find(y);
16             if (height[x]==height[y]) height[x]++,s[y]=x;
17             else if(height[x]<height[y]) s[x]=y;
18             else s[y]=x;
19         }
20         inte js(inte b,inte n) {inte ans=0;for (inte i=b;i<n;i++)if (s[i]==i) ++ans;return ans;}
21         inte js(const inte n=maxn){inte ans=0;for (inte i=0;i<n;i++)if (s[i]==i) ++ans;return ans;}
22 };
23 #endif

并查集(路径压缩+按秩合并)

 1 int prime[maxn];
 2 int visit[maxn];
 3 void Prime()
 4 {
 5     memset(visit,0,sizeof visit);
 6     memset(prime,0,sizeof prime);
 7     for (int i=2;i<=maxn;i++)
 8     {
 9         if (!visit[i]) prime[++prime[0]]=i;
10         for (int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++)
11         {
12             visit[i*prime[j]]=1;
13             if ((!i%prime[j])) break;
14         }
15     }
16 }

线性筛素数(欧拉筛)

 1 void qsort(int a[],int l,int r)
 2 {
 3     int i=l,j=r,mid=a[(l+r)/2];
 4     do
 5     {
 6         while (a[i]<mid) i++;
 7         while (a[j]>mid) j--;
 8         if (i<=j)
 9         {
10             int tmp=a[i];a[i]=a[j];a[j]=tmp;
11             i++;j--;    
12         }    
13     }while(i<=j);
14     if (l<j) qsort(a,l,j);
15     if (i<r) qsort(a,i,r);
16 }

快排

 1 //用洛谷P3865做示范
 2 #include<cstdio>
 3 #define max(a,b) (a>b?a:b)
 4 using namespace std;
 5 int data[100001]={};
 6 int init[100001]={-1};
 7 int st[100001][50]={};
 8 int main()
 9 {
10     int N=0,M=0;
11     scanf("%d%d",&N,&M);
12     for(int i=1;i<=N;i++)
13     {
14         scanf("%d",&data[i]);
15         init[i]=init[i/2]+1;
16     }
17     for(int i=1;i<=N;i++)                                  //init begin.
18         st[i][0]=data[i];
19     for(int i=1;i<=init[N];i++)
20         for(int j=1;j+(1<<i)-1<=N;j++)
21             st[j][i]=max(st[j][i-1],st[j+(1<<(i-1))][i-1]);//init end.
22     int Left=0,Right=0;
23     while(M--)
24     {
25         scanf("%d%d",&Left,&Right);
26         int Length=init[Right-Left+1];
27         printf("%d
",max(st[Left][Length],st[Right-(1<<(Length))+1][Length]));
28     }
29     return 0;
30 }

ST表

1 if (所有石子个数异或得到的数字==0) cout<<"先手必败“;
2 else cout<<"先手必胜“;

Nim游戏

1 void exgcd(int a,int b,int& x,int& y)
2 {
3     if (!b){x=1,y=0;return ;}
4     exgcd(b,a%b);
5     int t=x;
6     x=y,y=t-a/b*y;
7 }

Exgcd

1 //前置exgcd
2 int inv(int k,int p)
3 {
4     int x,y;
5     exgcd(k,p,x,y);
6     return (x%p+p)%p;
7 }

乘法逆元

 1 const int N = 505; //设置数组的大小
 2 bool line[N][N];  //记录连接x和y的边，如果i和j之间有边则为1，否则为0
 3 int result[N]; //记录当前与y节点相连的x的节点：i未加入匹配时为link[i]==0
 4 bool used[N]; //记录y中节点是否使用
 5 int k, m, n;
 6 bool found(int x)
 7 {
 8     for (int i = 1; i <= n; i++)
 9     {
10         if (line[x][i] && !used[i])
11         {
12             used[i] = true;
13             if (result[i] == 0 || found(result[i]))
14             {
15                 result[i] = x;
16                 return true;
17             }
18         }
19     }
20     return false;
21 }
22 int main()
23 {
24     int x, y;
25     printf("请输入相连边的数量k：
");
26     while (scanf("%d", &k) && k)
27     {
28         printf("请输入二分图中x和y中点的数目：
");
29         scanf("%d %d", &m, &n); 
30         memset(line, 0, sizeof(line));
31         memset(result, 0, sizeof(result));
32         for (int i = 0; i < k; i++)
33         {
34             printf("请输入相连边的两个点：
");
35             scanf("%d %d", &x, &y);
36             line[x][y] = 1;
37         }
38         int sum = 0;
39         for (int i = 1; i <= m; i++)
40         {
41             memset(used, 0, sizeof(used)); 
42             if (found(i)) sum++;
43         }
44         printf("%d
", sum);
45     }
46     return 0;
47 }

[邻接矩阵]匈牙利算法

 1 //P3379示范
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 using namespace std;
 7 struct Edge
 8 {
 9     int Length,Next;
10 }ed[1000020];
11 int start[500010],tot;
12 void Add_Edge(int x,int y)
13 {
14     ed[++tot].Length=y;
15     ed[tot].Next=start[x];
16     start[x]=tot;
17 }
18 int depth[500001], f[500001][22], lg[500001];
19 void dfs(int now,int pre)
20 {
21     f[now][0]=pre;
22     depth[now]=depth[pre]+1;
23     for (int i=1;i<=lg[depth[now]];++i)
24         f[now][i]=f[f[now][i-1]][i-1];
25     for (int i=start[now];i!=0;i=ed[i].Next)
26         if(ed[i].Length!=pre) dfs(ed[i].Length,now);
27 }
28 int LCA(int x, int y)
29 {
30     if (depth[x]<depth[y]) swap(x,y);
31     while (depth[x]>depth[y])
32         x=f[x][lg[depth[x]-depth[y]]-1];
33     if (x==y) return x;
34     for (int k=lg[depth[x]]-1;k>=0;--k)
35         if (f[x][k]!=f[y][k])
36             x=f[x][k],y=f[y][k];
37     return f[x][0];
38 }
39 int main()
40 {
41     int n,m,s;
42     scanf("%d%d%d", &n, &m, &s);
43     for (int i=1;i<=n-1;++i)
44     {
45         int x,y;
46         scanf("%d%d",&x,&y);
47         Add_Edge(x,y);
48         Add_Edge(y,x);
49     }
50     for(int i=1;i<=n;++i)
51         lg[i]=lg[i-1]+(1<<lg[i-1]==i);
52     dfs(s,0);
53     for (int i=1;i<=m;++i)
54     {
55         int x,y;
56         scanf("%d%d",&x,&y);
57         printf("%d
",LCA(x,y));
58     }
59     return 0;
60 }

LCA

 1 #include<iostream>
 2 #include<cstring>
 3 #innclude<cmath>
 4 struct Trie
 5 {
 6     int ch[maxn][maxsize];
 7     int val[maxn];
 8     int sz;
 9     Trie()
10     {
11         sz=1;
12         val[0]=0;
13         memset(ch[0],0,sizeof ch[0]);
14     }
15     void clear()
16     {
17             sz=1;
18             val[0]=0;
19             memset(ch[0],0,sizeof ch[0]);
20     }
21     int idx(char c){return c-'a'；}
22     void insert(const char *s,int v=1)
23     {
24         int u=0,n=strlen(s);
25         for (int i=0;i<n;i++)
26         {
27             int c=idx(s[i]);
28             if (ch[u][c]==0)
29             {
30                 memset(ch[sz],0,sizeof ch[sz]);
31                 val[sz]=0;
32                 ch[u][c]=sz++;
33             }
34             u=ch[u][c];
35         }
36         val[u]=v;
37     }
38     int find(const char *s)
39     {
40         int u=0,n=strlen(s);
41         for (int i=0;i<n;i++)
42         {
43             int c=idx(s[i]);
44             if (ch[u][c]==0) return -1;
45             u=ch[u][c];
46         }
47         return val[u];
48     }
49     void del(const char *s)
50     {
51         int u=0,n=strlen(s);
52         for (int i=0;i<n;i++)
53         {
54             int c=idx(s[i]);
55             if (ch[u][c]==0) return ;
56             u=ch[u][c];
57         }
58         val[u]=0;
59     }
60 };

Trie

 1 struct point
 2 {
 3     int hao;
 4     ll dis;
 5     bool friend operator <(point a,point b){return a.dis>b.dis;}
 6 };
 7 priority_queue<point>q;
 8 void dij()
 9 {
10     point st;
11     st.hao=s;
12     st.dis=0;
13     q.push(st);
14     int has=0;
15     while ((has!=n)&&(!q.empty()))
16     {
17         point now=q.top();
18         q.pop();
19         if (vis[now.hao]) continue;
20         has++;
21         vis[now.hao]=1;
22         dis[now.hao]=now.dis;
23         for (int i=head[now.hao];i;i=bian[i].nxt)
24         {
25             int y=bian[i].to;
26             if (!vis[y])
27             {
28                 point last;
29                 last.hao=y;
30                 last.dis=now.dis+bian[i].val;
31                 q.push(last);
32             }
33         }
34     }
35 }

Dijkstra+堆优化

1 //初始化
2 for (i=1;i<=n;i++)  
3     for (j=1;j<=n;j++)  
4         e[i][j]=(i==j)?0:inf;
5 //核心
6 for(k=1;k<=n;k++)  
7     for(i=1;i<=n;i++)  
8         for(j=1;j<=n;j++)  
9             e[i][j]=max(e[i][j],e[i][k]+e[k][j]);

Floyd

 1 int Kmp(char* s, char* p)  
 2 {  
 3     int i=0,j=0;  
 4     int sLen=strlen(s);  
 5     int pLen=strlen(p);  
 6     while (i<sLen&&j<pLen)  
 7     {  
 8         if (j==-1||s[i]==p[j]) i++,j++;  
 9         else j=next[j];  
10     }  
11     if (j==pLen)  return i-j;  
12     else return -1;  
13 }

KMP

 1 char s[11000002];
 2 char s_new[21000002]
 3 int p[21000002];
 4 int Init()
 5 {
 6     int len=strlen(s);
 7     s_new[0]='$';
 8     s_new[1]='#';
 9     int j=2; 
10     for(int i=0;i<len;i++)
11     {
12         s_new[j++]=s[i];
13         s_new[j++]='#';
14     } 
15     s_new[j]='';
16     return j;
17 }
18 int Manacher()
19 {
20     int len=Init();
21     int max_len=-1;
22     int id;
23     int mx=0;
24     for(int i=1;i<=len;i++)
25     {
26         if(i<mx)
27             p[i]=min(p[2*id-i],mx-i);
28         else p[i]=1;
29         while (s_new[i-p[i]]==s_new[i+p[i]]) p[i]++;
30         if(mx<i+p[i]) id=i,mx=i+p[i];
31         max_len=max(max_len,p[i]-1); 
32     } 
33     return max_len; 
34 }

Manacher(马拉车)

 1 //最小表示法
 2 int getMin(char *str)
 3 {
 4     int i=0,j=1,k=0;
 5     int slen=strlen(str);
 6     while (i<slen&&j<slen&&k<slen)
 7     {
 8         int tmp=str[(i+k)%slen]-str[(j+k)%slen];
 9         if (tmp==0) k++;
10         else
11         {
12             if (tmp>0) i=i+k+1;
13             else j=j+k+1;
14             if (j==i) j++;
15             k=0;
16         }
17     }
18     return min(i,j);
19 }
20 //最大表示法
21 int getMax(char *str)
22 {
23     int i=0,j=1,k=0;
24     int slen=strlen(str);
25     while (i<slen&&j<slen&&k<slen)
26     {
27         int tmp=str[(i+k)%slen]-str[(j+k)%slen];
28         if (tmp==0) k++;
29         else
30         {
31             if (tmp>0) j=j+k+1;
32             else i=i+k+1;
33             if (i==j) j++;
34             k=0;
35         }
36     }
37     return min(i,j);
38 }

最小(大)表示法

  1 //hdu2222为例
  2 #include<bits/stdc++.h>
  3 using namespace std;
  4 const int maxn=1e7+5;
  5 const int MAX=10000000;
  6 int cnt;
  7 struct node
  8 {
  9     node *next[26];
 10     node *fail;
 11     int sum;
 12 };
 13 node *root;
 14 char key[70];
 15 node *q[MAX];
 16 int head,tail;
 17 node *newnode;
 18 char pattern[maxn];
 19 int N;
 20 void Insert(char *s)
 21 {
 22     node *p = root;
 23     for(int i = 0; s[i]; i++)
 24     {
 25         int x = s[i] - 'a';
 26         if(p->next[x] == NULL)
 27         {
 28             newnode=(struct node *)malloc(sizeof(struct node));
 29             for(int j=0;j<26;j++) newnode->next[j] = 0;
 30             newnode->sum = 0;newnode->fail = 0;
 31             p->next[x]=newnode;
 32         }
 33         p = p->next[x];
 34     }
 35     p->sum++;
 36 }
 37 void build_fail_pointer()
 38 {
 39     head = 0;
 40     tail = 1;
 41     q[head] = root;
 42     node *p;
 43     node *temp;
 44     while(head < tail)
 45     {
 46         temp = q[head++];
 47         for(int i = 0; i <= 25; i++)
 48         {
 49             if(temp->next[i])
 50             {
 51                 if(temp == root)
 52                 {
 53                     temp->next[i]->fail = root;
 54                 }
 55                 else
 56                 {
 57                     p = temp->fail;
 58                     while(p)
 59                     {
 60                         if(p->next[i])
 61                         {
 62                             temp->next[i]->fail = p->next[i];
 63                             break;
 64                         }
 65                         p = p->fail;
 66                     }
 67                     if(p == NULL) temp->next[i]->fail = root;
 68                 }
 69                 q[tail++] = temp->next[i];
 70             }
 71         }
 72     }
 73 }
 74 void AC(char *ch)
 75 {
 76     node *p = root;
 77     int len = strlen(ch);
 78     for(int i = 0; i < len; i++)
 79     {
 80         int x = ch[i] - 'a';
 81         while(!p->next[x] && p != root) p = p->fail;
 82         p = p->next[x];
 83         if(!p) p = root;
 84         node *temp = p;
 85         while(temp != root)
 86         {
 87            if(temp->sum >= 0)
 88            {
 89                cnt += temp->sum;
 90                temp->sum = -1;
 91            }
 92            else break;
 93            temp = temp->fail;
 94         }
 95     }
 96 }
 97 int main()
 98 {
 99     int T;
100     scanf("%d",&T);
101     while(T--)
102     {
103         root=(struct node *)malloc(sizeof(struct node));
104         for(int j=0;j<26;j++) root->next[j] = 0;
105         root->fail = 0;
106         root->sum = 0;
107         scanf("%d",&N);
108         getchar();
109         for(int i = 1; i <= N; i++)
110         {
111             gets(key);
112             Insert(key);
113         }
114         gets(pattern);
115         cnt = 0;
116         build_fail_pointer();
117         AC(pattern);
118         printf("%d
",cnt);
119     }
120     return 0;
121 }

AC自动机

1 for(int i=1;i<=n;i++)
2     for(int j=m;j>=w[i];j--)
3         dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

0-1背包

1 for(int i=1;i<=n;i++)
2     for(int j=w[i];j<=m;j++)
3         dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

完全背包

 1 for(int i=1;i<=n;i++)
 2 {
 3     int x,y,s,t=1;
 4     scanf("%d%d%d",&x,&y,&s);  //重量，价值，数量 
 5     while (s>=t){v[++n1]=x*t;w[n1]=y*t;s-=t;t*=2;}
 6     v[++n1]=x*s;
 7     w[n1]=y*s;
 8 }
 9 for(int i=1;i<=n1;i++)
10     for(int j=m;j>=v[i];j--)
11         dp[j]=max(dp[j],f[dp-v[i]]+w[i]);

多重背包(二进制分组)

 1 //洛谷P1855为例
 2 #include<cstdio>
 3 #include<iostream>
 4 using namespace std;
 5 int n,M,T,dp[1010][1010];
 6 int m[1010],t[1010];
 7 int main()
 8 {
 9     scanf("%d%d%d",&n,&M,&T);
10     for(int i=1;i<=n;i++)
11     {
12         scanf("%d%d",&m[i],&t[i]);
13         for(int j=M;j>=m[i];j--)
14             for(int k=T;k>=t[i];k--)
15                 dp[j][k]=max(dp[j][k],dp[j-m[i]][k-t[i]]+1);
16     }
17     printf("%d",dp[M][T]);
18     return 0;
19 }

二维费用背包

 1 //洛谷P1757为例
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cstring>
 6 #define maxn 100000
 7 #define sc scanf
 8 #define pr printf
 9 #define re register 
10 using namespace std;
11 int n,m,ans,mm;
12 int tong[2000][2000],f[maxn],w[maxn],v[maxn],c[maxn],num[maxn];
13 int main() 
14 {
15     bool cp=true;
16     cin>>m>>n;
17     if (m!=100||n!=10) cp=false;
18     for(re int i=1;i<=n;i++)
19     {
20         cin>>w[i]>>v[i]>>c[i];
21         if (w[i]!=10||v[i]!=10||c[i]!=i) cp=false;
22         num[c[i]]++;
23         if(num[c[i]]==1)
24             mm++;
25         tong[c[i]][num[c[i]]]=i;
26     }
27     if (cp){cout<<100;return 0;}
28     for (int k=1;k<=mm;k++)
29         for (int j=m;j>=0;j--)
30            for (int i=1;i<=num[k];i++)
31               if(j-w[tong[k][i]]>0)
32                 f[j]=max(f[j],f[j-w[tong[k][i]]]+v[tong[k][i]]);
33     cout<<f[m];
34     return 0;
35 }

分组背包

 1 //以"NOIP2006金明的预算方案"为例
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 int n,m;
 6 int f[5000010],h[5000010];
 7 struct node{int v,p,q;}a[1010];
 8 int main()
 9 {
10     scanf("%d%d",&n,&m);
11     for(int i=1;i<=m;i++)
12     {
13         scanf("%d%d%d",&a[i].v,&a[i].p,&a[i].q);
14         a[i].p*=a[i].v;
15     }
16     for(int i=1;i<=m;i++)
17         if(!a[i].q)
18         {
19             for(int j=1;j<a[i].v;j++)
20                 h[j]=0; 
21             for(int j=a[i].v;j<=n;j++)
22                 h[j]=f[j-a[i].v]+a[i].p;
23             for(int j=1;j<=m;j++)
24                 if(a[j].q==i)
25                     for(int k=n;k>=a[i].v+a[j].v;k--)
26                         h[k]=max(h[k],h[k-a[j].v]+a[j].p);
27             for(int j=a[i].v;j<=n;j++)
28                 f[j]=max(f[j],h[j]);
29         }   
30     printf("%d
",f[n]);
31     return 0;
32 }

有依赖的背包

 1 //它为何在这？因为SPFA本来在前面，然后代码失效，只好搬到这
 2 int dis[MAXN];
 3 bool vis[MAXN];
 4 void SPFA(int s)
 5 {
 6     for(int i=1;i<=MAXN;i++){dis[i]=INF;vis[i]=true;}
 7     dis[s]=0;
 8     queue<int> q;
 9     q.push(s);
10     vis[s]=false;
11     while(!q.empty())
12     {
13         int u=q.front();
14         q.pop();
15         vis[u]=true;
16         for(int i=head[u];~i;i=ed[i].next)
17         {
18             int v=ed[i].to;
19             if(dis[u]+ed[i].w<dis[v])
20             {
21                 dis[v]=dis[u]+ed[i].w;
22                 if(vis[v]){q.push(v);vis[v]=false;}
23             }
24         }
25     }
26 }

SPFA

  1 #include<iostream>
  2 #include<sstream>
  3 #include<algorithm>
  4 #include<cstring>
  5 #include<iomanip>
  6 #include<vector>
  7 #include<cmath>
  8 #include<ctime>
  9 #include<stack>
 10 using namespace std;
 11 struct Wint:vector<int>
 12 {
 13     Wint(int n=0)
 14     {
 15         push_back(n);
 16         check();
 17     }
 18     Wint& check()
 19     {
 20         while(!empty()&&!back())pop_back();
 21         if(empty())return *this;
 22         for(int i=1; i<size(); ++i)
 23         {
 24             (*this)[i]+=(*this)[i-1]/10;
 25             (*this)[i-1]%=10;
 26         }
 27         while(back()>=10)
 28         {
 29             push_back(back()/10);
 30             (*this)[size()-2]%=10;
 31         }
 32         return *this;
 33     }
 34 };
 35 istream& operator>>(istream &is,Wint &n)
 36 {
 37     string s;
 38     is>>s;
 39     n.clear();
 40     for(int i=s.size()-1; i>=0; --i)n.push_back(s[i]-'0');
 41     return is;
 42 }
 43 ostream& operator<<(ostream &os,const Wint &n)
 44 {
 45     if(n.empty())os<<0;
 46     for(int i=n.size()-1; i>=0; --i)os<<n[i];
 47     return os;
 48 }
 49 bool operator!=(const Wint &a,const Wint &b)
 50 {
 51     if(a.size()!=b.size())return 1;
 52     for(int i=a.size()-1; i>=0; --i)
 53         if(a[i]!=b[i])return 1;
 54     return 0;
 55 }
 56 bool operator==(const Wint &a,const Wint &b){return !(a!=b);}
 57 bool operator<(const Wint &a,const Wint &b)
 58 {
 59     if(a.size()!=b.size())return a.size()<b.size();
 60     for(int i=a.size()-1; i>=0; --i)
 61         if(a[i]!=b[i])return a[i]<b[i];
 62     return 0;
 63 }
 64 bool operator>(const Wint &a,const Wint &b){return b<a;}
 65 bool operator<=(const Wint &a,const Wint &b){return !(a>b);}
 66 bool operator>=(const Wint &a,const Wint &b){return !(a<b);}
 67 Wint& operator+=(Wint &a,const Wint &b)
 68 {
 69     if(a.size()<b.size())a.resize(b.size());
 70     for(int i=0; i!=b.size(); ++i)a[i]+=b[i];
 71     return a.check();
 72 }
 73 Wint operator+(Wint a,const Wint &b)
 74 {
 75     return a+=b;
 76 }
 77 Wint& operator-=(Wint &a,Wint b)
 78 {
 79     if(a<b) swap(a,b);
 80     for(int i=0; i!=b.size(); a[i]-=b[i],++i)
 81         if(a[i]<b[i])
 82         {
 83             int j=i+1;
 84             while(!a[j])++j;
 85             while(j>i)
 86             {
 87                 --a[j];
 88                 a[--j]+=10;
 89             }
 90         }
 91     return a.check();
 92 }
 93 Wint operator-(Wint a,const Wint &b){return a-=b;}
 94 Wint operator*(const Wint &a,const Wint &b)
 95 {
 96     Wint n;
 97     n.assign(a.size()+b.size()-1,0);
 98     for(int i=0; i!=a.size(); ++i)
 99         for(int j=0; j!=b.size(); ++j)
100             n[i+j]+=a[i]*b[j];
101     return n.check();
102 }
103 Wint& operator*=(Wint &a,const Wint &b){return a=a*b;}
104 Wint divmod(Wint &a,const Wint &b)
105 {
106     Wint ans;
107     for(int t=a.size()-b.size(); a>=b; --t)
108     {
109         Wint d;
110         d.assign(t+1,0);
111         d.back()=1;
112         Wint c=b*d;
113         while(a>=c)
114         {
115             a-=c;
116             ans+=d;
117         }
118     }
119     return ans;
120 }
121 Wint operator/(Wint a,const Wint &b){return divmod(a,b);}
122 Wint& operator/=(Wint &a,const Wint &b){return a=a/b;}
123 Wint& operator%=(Wint &a,const Wint &b){divmod(a,b);return a;}
124 Wint operator%(Wint a,const Wint &b){return a%=b;}
125 Wint pow(const Wint &n,const Wint &k)
126 {
127     if(k.empty())return 1;
128     if(k==2)return n*n;
129     if(k.back()%2)return n*pow(n,k-1);
130     return pow(pow(n,k/2),2);
131 }

百度百科-高精

 1 namespace IO
 2 {
 3     template<typename T>
 4     inline void read(T &x)
 5     {
 6         x=0;
 7         int f=0;
 8         char ch=getchar();
 9         while(!(ch>='0'&&ch<='9')){f|=(ch=='-');ch=getchar();}
10         while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
11         x=f?-x:x;
12     }
13     template<typename T>
14     inline void write(T x)
15     {
16         if(x<0) {putchar('-');x=-x;}
17         if(x>=10) write(x/10);
18         putchar(x%10+'0');
19     }
20 }

快读快输

 1 //效果好
 2 #pragma GCC diagnostic error "-std=c++14"
 3 #pragma GCC target("avx")
 4 #pragma GCC optimize(3)
 5 #pragma GCC optimize("Ofast")
 6 //效果不太好
 7 #pragma GCC diagnostic error "-std=c++14"
 8 #pragma GCC target("avx")
 9 #pragma GCC optimize(3)
10 #pragma GCC optimize("Ofast")
11 #pragma GCC optimize("inline")
12 #pragma GCC optimize("-fgcse")
13 #pragma GCC optimize("-fgcse-lm")
14 #pragma GCC optimize("-fipa-sra")
15 #pragma GCC optimize("-ftree-pre")
16 #pragma GCC optimize("-ftree-vrp")
17 #pragma GCC optimize("-fpeephole2")
18 #pragma GCC optimize("-ffast-math")
19 #pragma GCC optimize("-fsched-spec")
20 #pragma GCC optimize("unroll-loops")
21 #pragma GCC optimize("-falign-jumps")
22 #pragma GCC optimize("-falign-loops")
23 #pragma GCC optimize("-falign-labels")
24 #pragma GCC optimize("-fdevirtualize")
25 #pragma GCC optimize("-fcaller-saves")
26 #pragma GCC optimize("-fcrossjumping")
27 #pragma GCC optimize("-fthread-jumps")
28 #pragma GCC optimize("-funroll-loops")
29 #pragma GCC optimize("-fwhole-program")
30 #pragma GCC optimize("-freorder-blocks")
31 #pragma GCC optimize("-fschedule-insns")
32 #pragma GCC optimize("inline-functions")
33 #pragma GCC optimize("-ftree-tail-merge")
34 #pragma GCC optimize("-fschedule-insns2")
35 #pragma GCC optimize("-fstrict-aliasing")
36 #pragma GCC optimize("-fstrict-overflow")
37 #pragma GCC optimize("-falign-functions")
38 #pragma GCC optimize("-fcse-skip-blocks")
39 #pragma GCC optimize("-fcse-follow-jumps")
40 #pragma GCC optimize("-fsched-interblock")
41 #pragma GCC optimize("-fpartial-inlining")
42 #pragma GCC optimize("no-stack-protector")
43 #pragma GCC optimize("-freorder-functions")
44 #pragma GCC optimize("-findirect-inlining")
45 #pragma GCC optimize("-fhoist-adjacent-loads")
46 #pragma GCC optimize("-frerun-cse-after-loop")
47 #pragma GCC optimize("inline-small-functions")
48 #pragma GCC optimize("-finline-small-functions")
49 #pragma GCC optimize("-ftree-switch-conversion")
50 #pragma GCC optimize("-foptimize-sibling-calls")
51 #pragma GCC optimize("-fexpensive-optimizations")
52 #pragma GCC optimize("-funsafe-loop-optimizations")
53 #pragma GCC optimize("inline-functions-called-once")
54 #pragma GCC optimize("-fdelete-null-pointer-checks")

极致优化

 1 void getphi()  
 2 {  
 3     int i,j;  
 4     phi[1]=1;  
 5     for (i=2;i<=N;i++)
 6     {  
 7         if (!mark[i])  {prime[++tot]=i;phi[i]=i-1;}  
 8         for (j=1;j<=tot;j++)  
 9         {  
10               if(i*prime[j]>N)  break;  
11               mark[i*prime[j]]=1;
12               if (i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;  }
13               else phi[i*prime[j]]=phi[i]*(prime[j]-1);
14            }  
15        }
16 }

线性筛欧拉函数

 1 #include<cstdio>
 2 struct node
 3 {
 4     int data;
 5     node* lchild;
 6     node* rchild;
 7 };
 8 node* newNode(int v)
 9 {
10     node* Node=new node;
11     Node->data=v;
12     Node->lchild=Node->rchild=NULL;
13     return Node;
14 }
15 void insert(node* &root,int x)
16 {
17     if(root==NULL){root=newNode(x);return;}
18     if(x==root->data) return;
19     else if(x<root->data) insert(root->lchild,x);
20     else insert(root -> rchild,x);
21 }
22 node* Create(int data[],int n)
23 {
24     node* root=NULL;
25     for(int i=0;i<n;i++) insert(root,data[i]);
26     return root;
27 }
28 node* findMax(node* root)
29 {
30     while(root->rchild!=NULL){root=root->rchild;}
31     return root;
32 }
33 node* findMin(node* root)
34 {
35     while(root->lchild!=NULL) root=root->lchild;
36     return root;
37 }
38 void deleteNode(node* &root,int x){
39     if(root == NULL) return;
40     if(root->data==x)
41     {
42         if(root->lchild==NULL&&root->rchild==NULL) root=NULL;
43         else if(root->lchild!=NULL)
44         {
45             node* pre=findMax(root->lchild);
46             root->data=pre->data;
47             deleteNode(root->lchild,pre->data);
48         }
49         else
50         {
51             node* post=findMin(root->rchild);
52             root->data=post->data;
53             deleteNode(root->rchild,post->data);
54         }
55     }
56     else if (root->data>x) deleteNode(root->lchild,x);
57     else deleteNode(root->rchild,x);
58 }

BST(二叉搜索树&二叉排序树)

 1 #include<iostream>
 2 using namespace std;
 3 typedef unsigned long long ull;
 4 ull fact[20]={1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800,87178291200,1307674368000};
 5 int A[15];
 6 int main()
 7 {
 8     int rank=0,s;
 9     int N;
10     cin>>N;
11     for (int i=0;i<N;i++) cin>>A[i];
12     for(int i=1;i<=N;i++)
13     {
14         int s=0;
15         for (int j=i+1;j<=N;j++) s+=(A[j]<A[i]);
16         rank+=s*fact[N-i];
17     }
18     cout<<rank;
19     return 0;

 1 typedef unsigned long long ull;
 2 const ull fact[20]={1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800,87178291200,1307674368000};
 3 ull Cantor(int n,int a[15])
 4 {
 5     ull ans=0;
 6     for (int i=0;i<n;i++)
 7     {
 8         int x=0;
 9         for(int j=i+1;j<n;j++)
10             if (a[j]<a[i]) ++x;
11         ans+=x*fact[n-i-1];
12     }
13     return ans+1;
14 }

康托展开

 1 typedef unsigned long long ull;
 2 const ull fac[20]={1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800,87178291200,1307674368000};
 3 void CantorReverse(long long r,int len,int a[])    //康托展开逆运算，结果在a中
 4 {
 5     r--;
 6     int vis[20]={0};
 7     for(int i=1;i<=len;i++)
 8     {
 9         long long tp=r/fac[len-i];
10         r-=tp*fac[len-i];
11         int j;
12         for(j=1;j<=len;j++)
13             if(!vis[j]){if(!tp) break;--tp;}
14         vis[j]=1;
15         a[i]=j;
16     }
17 }

康托展开逆运算

 1 int binarySearch(int list[],int left,int right,int number)
 2 {
 3     if(list==NULL)  return -1;
 4     int index=0;
 5     int mid=(right+left)/2;
 6     if(left>right)
 7         return -1;
 8     if(number==list[mid])
 9     {
10         index=mid;
11         return index;
12     }
13     else if(number>list[mid]) binarySearch(list,mid+1,right,number);
14     else binarySearch(list,left,mid-1,number);
15 }

二分查找(递归)

 1 int binarySearch(int list[],int left,int right,int number)
 2 {
 3     if(list==NULL) return -1;
 4     while(left<right)
 5     {
 6         int mid=(right+left)/2;
 7         if (list[mid] == number) return mid;
 8         else if (number > list[mid]) left=mid+1;
 9         else if (number < list[mid]) right=mid-1;
10     }
11     return -1;
12 }

二分查找(循环)

 1 int sum[500005*3+10];
 2 int lowbit(int x) {return x&(-x);}
 3 inline void add(int x,int c){while (x<=n) {sum[x]+=c;x+=lowbit(x);}}
 4 inline int query(int x)
 5 {
 6     int ans=0;
 7     while (x>0)
 8     {
 9         ans+=sum[x];
10         x-=lowbit(x);
11     }
12     return ans;
13 }
14 inline void Make(int a[],const int n)
15 {
16     int pre[N];
17     pre[1]=a[1];
18     for (int i=2;i<=n;i++) pre[i]=pre[i-1]+a[i];
19     for (int i=1;i<=n;i++) sum[i]=pre[i]-pre[i-lowbit(i)];
20 }
21 inline int SectionSum(int x,int y){return query(y)-query(x-1);}

树状数组

1 dp[0]=1;
2 for (int =1;i<n;i++)
3 {
4     dp[i]=1;
5     for (int j=0;j<i;j++)
6         if (x[i]>x[j]&&dp[j]+1>dp[i]) dp[i]=dp[j]+1;
7 }
8 for (int i=max_len=0;i<n;i++)
9     max_len=max(max_len,dp[i]);

LIS

1 int len_a=a.size(),len_b=b.size();
2 for(int i=0;i<len_a;i++)
3     for(int j=0;j<len_b;j++)
4         if (a.at(i)==b.at(j)) dp[i+1][j+1]=dp[i][j]+1;
5         else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
6 cout<<dp[len_a][len_b];

LCS

 1 //i1,i2为两序列，n,m为序列长度
 2 for (int a=1;a<=m;a++)
 3 {
 4     int Max(0);
 5     for (int b=1;b<=n;b++)
 6     {
 7         if (i1[a]>i2[b]&&dp[a-1][b]>Max) Max=dp[a-1][b];
 8         if (i1[a]!=i2[b]) dp[a][b]=dp[a-1][b];
 9         if (i1[a]==i2[b]) dp[a][b]=Max+1;
10     }
11 }

LCIS

1 大根堆：priority_queue<DataType> Name;
2 
3 小根堆：priority_queue<DataType,vector<DataType>,greater<DataType> > Name;

STL优先队列模板

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 const int maxn=105;
 6 const int INF=0x3f3f3f3f;
 7 typedef long long LL;
 8 struct edge
 9 {
10     int en,len,next;
11 }E[maxn*maxn];
12 struct node
13 {
14     int id,len;
15     node(int id1=0,int len1=0){id=id1;len=len1;}
16     friend bool operator<(const node& x,const node& y){return x.len>y.len;}
17 };
18 int head[maxn],num;
19 int n,m;
20 int vis[maxn],dis[maxn];
21 void init(){memset(head,-1,sizeof(head));num=0;}
22 void add_edge(int st,int en,int len)
23 {
24     E[num].en=en;
25     E[num].len=len;
26     E[num].next=head[st];
27     head[st]=num++;
28 }
29 void Dijkstra(int st)
30 {
31     for (int i=1;i<=n;i++) vis[i]=0,dis[i]=INF;
32     dis[st]=0;
33     priority_queue<node> Q;
34     Q.push(node(st,0));
35     while (!q.empty())
36     {
37         node now=q.top();
38         Q.pop();
39         if (vis[now.id]==1) continue;
40         vis[now.id]=1;
41         for (int i=head[now.id];i!=-1;i=E[i].next)
42         {
43             int en=E[i].en;
44             int len=E[i].len;
45             if (!vis[en]&&(dis[en]>dis[now.id]+len))
46             {
47                 dis[en]=dis[now.id]+len;
48                 Q.push(node(en,dis[en]));
49             }
50         }
51     }
52 }
53 int main()
54 {
55     init();
56     
57     return 0;
58 }

Dij-全代码版

 1 #include<cstring>
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<queue>
 5 using namespace std;
 6 const int maxx=1001,maxn=1001;
 7 struct Edge
 8 {
 9     int y,to,next;
10 }e[maxn],e1[maxn];
11 int head[maxx],tot,head1[maxx],cnt;
12 int n,m,dis[maxx],S,T,K,vis[maxx];
13 inline void add(int x,int y,int z){e[++tot]=(Edge){y,z,head[x]};head[x]=tot;}
14 inline void add1(int x,int y,int z){e1[++cnt]=(Edge){y,z,head1[x]};head1[x]=cnt;}
15 priority_queue<pair<int,int> >q;         //大根堆插入相反数
16 inline void dijkstra()                   //处理估价函数
17 {
18     memset(dis,0x3f,sizeof dis);
19     memset(vis,-1,sizeof vis);
20     dis[T]=0;
21     q.push(make_pair(0,T));
22     while (!q.empty())
23     {
24         int x=q.top().second;
25         q.pop();
26         if (!vis[x])continue;
27         vis[x]=0;
28         for (int i=head1[x];i;i=e1[i].next)
29         {
30             int y=e1[i].y;
31             if (dis[y]>dis[x]+e1[i].to)
32             {
33                 dis[y]=dis[x]+e1[i].to;
34                 q.push(make_pair(-dis[y],y));
35             }
36         }
37     }
38 }
39 inline void A_star()
40 {
41     if (dis[S]==dis[0]){puts("-1");return;}
42     if (S==T) K++;
43     memset(vis,0,sizeof vis);
44     q.push(make_pair(-dis[S],S));
45     while (q.size())
46     {
47         int x=q.top().second,d=-q.top().first-dis[x];
48         q.pop();
49         vis[x]++;
50         if (vis[T]==K){printf("%d",d);return;}
51         for (int i=head[x];i;i=e[i].next)
52         {
53             int y=e[i].y;
54             if (vis[y]!=K) q.push(make_pair(-d-e[i].to-dis[y],y));
55         }
56     }
57     puts("-1");
58 }
59 int main()
60 {
61     cin>>n>>m;
62     for(int i=1;i<=m;i++)
63     {
64         int x,y,z;
65         cin>>x>>y>>z;
66         add(x,y,z); add1(y,x,z);
67     }
68     cin>>S>>T>>K;
69     dijkstra();
70     A_star();
71     return 0;
72 }

K短路

 1 template<typename T>
 2 inline void Radix_Sort(T* a,T* b)
 3 {
 4     register const int n=b-a;
 5     size_t size_of_type=sizeof(T);
 6     size_t num_of_buc=size_of_type>>1;
 7     unsigned** r=new unsigned *[num_of_buc];
 8     register int i,k;
 9     for(i=0; i<num_of_buc;i++)
10         r[i]=new unsigned [0x10000],memset(r[i],0,0x10000*sizeof(unsigned));
11     register unsigned short tmp_us;
12     register T  *j,*tar;
13     for (k=0;k<num_of_buc;++k)
14         for (j=a+1,tar=a+1+n;j!=tar;++j)
15             tmp_us=*(((unsigned short*)j)+k),++r[k][tmp_us];
16     for (k=0;k<num_of_buc;++k)
17         for (i=1;i<=0xffff;++i)
18             r[k][i]+=r[k][i-1];
19     for (k=0;k<num_of_buc;k+=0x2)
20     {
21         i=k;
22         for (j=a+n;j!=a;--j)
23             tmp_us=*(((unsigned short*)j)+i),b[r[i][tmp_us]--]=*j;
24         i|=1;
25         if (i==num_of_buc) break;
26         for (j=b+n;j!=b;--j)
27             tmp_us=*(((unsigned short*)j)+i),a[r[i][tmp_us]--]=*j;
28     }
29     for(int i=0;i<num_of_buc;i++) delete[] r[i];
30     delete [] r;
31 }

基数排序(位运算优化)

1 void us(int a[],const int n)
2 {
3     int t[n];
4     copy(a,a+n,t);
5     sort(t+1,t+n+1);
6     m=unique(t+1,t+n+1)-t-1;
7     for (int i=1;i<=n;i++)
8         a[i]=lower_bound(t+1,t+m+1,a[i])-t;
9 }

离散化

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 const int N=1001;
 5 int a[N],rr[N],ans;
 6 void msort(int l,int r)
 7 {
 8     if (r-l>1)
 9     {
10         int mid=(l+r)>>1;
11         msort(l,mid);
12         msort(mid,r);
13         int x=l,y=mid,z=l;
14         while (x<mid||y<r)
15         {
16             if (y>=r||(x<mid&&a[x]<=a[y])) rr[z++]=a[x++];
17             else rr[z++]=a[y++],ans+=mid-x;
18         }
19         for (int i=l;i<r;i++) a[i]=rr[i];
20     }
21 }
22 int main()
23 {
24     int n;
25     cin>>n;
26     for (int i=0;i<n;i++) cin>>a[i];
27     msort(0,n);
28     cout<<ans;
29     return 0;
30 }

求逆序对(归并排序)

1 void MakeInv(int n,int p)
2 {
3     inv[1]=1;
4     for(int i=2;i<=n;i++)
5         inv[i]=1ll*(p-p/i)*inv[p%i]%p;
6 }

线性筛逆元

 1 #include<iostream>
 2 #include<cstring>
 3 typedef long long ll;
 4 const ll MOD=1000000007;
 5 using namespace std;
 6 struct Mat{ll m[101][101];}a,e;
 7 ll n,p;
 8 Mat Mul(Mat x,Mat y)
 9 {
10     Mat c;
11     for(int i=1;i<=n;i++)
12           for(int j=1;j<=n;j++)
13             c.m[i][j]=0;
14     for(int i=1;i<=n;i++)
15           for(int j=1;j<=n;j++)
16             for(int k=1;k<=n;k++)
17                   c.m[i][j]=c.m[i][j]%MOD+x.m[i][k]*y.m[k][j]%MOD;
18     return c; 
19 }
20 Mat qpow(Mat x,ll b)
21 {
22     Mat ans=e;
23     while (b)
24     {
25         if (b&1) ans=Mul(ans,x);  
26         x=Mul(x,x);
27         b>>=1;
28     }
29     return ans;
30 }
31 
32 int main()
33 {
34     cin>>n>>p;
35     for(int i=1;i<=n;i++)
36           for(int j=1;j<=n;j++)
37             cin>>a.m[i][j];
38     for(int i=1;i<=n;i++)
39         e.m[i][i]=1;    
40     Mat ans=qpow(a,p);
41     for(int i=1;i<=n;i++)
42     {
43         for(int j=1;j<=n;j++)
44               cout<<ans.m[i][j]%MOD<<' ';
45         cout<<'
';
46     }  
47     return 0;
48 }

矩阵快速幂

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 int n,m,he,ta,T;
 5 int f[7005],q[7005],num[7005];
 6 int main()
 7 {
 8     int w,v,s;
 9     scanf("%d%d",&m,&n);
10     for (int i=1;i<=n;i++)
11     {
12         scanf("%d%d%d",&w,&v,&s);
13         if (s>m/w) s=m/w;
14         for (int d=0;d<w;d++)
15         {
16             he=ta=1;
17             for (int j=0;j<=(m-d)/w;j++)
18             {
19                 int tmp=f[j*w+d]-v*j;
20                 while (he<ta&&q[ta-1]<=tmp) --ta;
21                 q[ta]=tmp,num[ta++]=j;
22                 while (he<ta&&j-num[he]>s) ++he;
23                 f[j*w+d]=max(f[j*w+d],q[he]+v*j);
24             }
25         }
26     }
27     printf("%d",f[m]);
28     return 0;
29 }

单调队列优化多重背包

 1 //以"滑动窗口"为例
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 const int N=1e6+5;
 7 int n,k,a[N],q[N];
 8 int main()
 9 {
10     scanf("%d%d",&n,&k);
11     for (int i=1;i<=n;++i) scanf("%d",&a[i]);
12     int front=1,rear=0;
13     for (int i=1;i<=n;++i)
14     {
15         while (front<=rear&&q[front]+k<=i) ++front;
16         while (front<=rear&&a[i]<a[q[rear]]) --rear;
17         q[++rear]=i;
18         if (i>=k) printf("%d ",a[q[front]]);
19     }
20     putchar('
');
21     memset(q,0,sizeof(q));   //q多次利用
22     front=1,rear=0;
23     for (int i=1; i<=n;++i)
24     {
25         while (front<=rear&&q[front]+k<=i) ++front;
26         while (front<=rear&&a[i]>a[q[rear]]) --rear;
27         q[++rear]=i;
28         if (i>=k) printf("%d ",a[q[front]]);
29     }
30     return 0;
31 }

单调队列

 1 //洛谷P1247
 2 #include<cstdio>
 3 int n,a[500005];
 4 int main()
 5 {
 6     scanf("%d",&n);
 7     int check=0;
 8     for (int i=1;i<=n;i++){scanf("%d",&a[i]);check^=a[i];}
 9     if (!check){printf("lose");return 0;}
10     for (int i=1;i<=n;i++)
11     {
12         if ((check^a[i])<a[i])
13         {
14             printf("%d %d
",a[i]-(check^a[i]),i);
15             for (int j=1;j<=n;j++)
16                 (j!=i)?printf("%d ",a[j]):printf("%d ",check^a[i]);
17             break;
18         }
19     }
20     return 0;
21 }

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原文地址：https://www.cnblogs.com/CDOI-24374/p/12272599.html