• 模板


    模板……有需要的人也可以拿走

    1、高精度模板(v1.0)

    注意:主要耗时在乘法、除法和乘方。其中高精度乘方有快速幂优化,常数大致是乘法的5~10倍。除法用二分写的效率较低,大概是乘法的20~30倍,能不用尽量不用

      1 #define mx 300
      2 struct gaojing{
      3      int len;
      4      int a[mx+10];
      5 }zero,one;
      6 inline void set0(gaojing &s)//高精清零
      7 {
      8     s.len=1;
      9     for (int i=1;i<=mx+5;i++)s.a[i]=0;
     10 }
     11 inline void inputn(gaojing &a)//高精输入 
     12 {
     13     set0(a);
     14     char ch=getchar();
     15     while (ch<'0'||ch>'9')ch=getchar();  
     16     while (ch>='0'&&ch<='9')  
     17     {  
     18         a.a[a.len++]=ch-'0';
     19         ch=getchar();
     20     }
     21     a.len--;  
     22     int change[mx+15];
     23     for (int i=1;i<=a.len;i++)  
     24       change[i]=a.a[i];  
     25     for (int i=1;i<=a.len;i++)  
     26       a.a[i]=change[a.len-i+1];
     27     while (a.a[a.len]==0)a.len--;
     28 }  
     29 inline void put(gaojing a)//高精输出
     30 {
     31     for (int i=a.len;i>=1;i--)printf("%d",a.a[i]);
     32     printf("
    ");
     33 }
     34 inline bool operator < (const gaojing &a,const gaojing &b)//高精< 
     35 {
     36     if (a.len<b.len)return 1;
     37     if (a.len>b.len)return 0;
     38     for (int i=a.len;i>=1;i--)
     39     {
     40         if (a.a[i]<b.a[i])return 1;
     41         if (a.a[i]>b.a[i])return 0;
     42     }
     43     return 0;
     44 }
     45 inline bool operator == (const gaojing &a,const gaojing &b)//高精==
     46 {
     47     if (a.len!=b.len)return 0;
     48     for (int i=a.len;i>=1;i--)
     49     {
     50         if (a.a[i]!=b.a[i])return 0;
     51     }
     52     return 1;
     53 }
     54 inline gaojing max(const gaojing &a,const gaojing &b)//高精max
     55 {
     56     if (a<b)return b;
     57     else return a;
     58 }
     59 inline gaojing min(const gaojing &a,const gaojing &b)//高精min
     60 {
     61     if (a<b)return a;
     62     else return b;
     63 } 
     64 inline gaojing operator + (const gaojing &a,const gaojing &b)//高精+
     65 {
     66     gaojing c;set0(c);  
     67     int maxlen=max(a.len,b.len);      
     68     for (int i=1;i<=maxlen;i++)      
     69     {      
     70         c.a[i]=c.a[i]+a.a[i]+b.a[i];      
     71         if (c.a[i]>=10)      
     72         {      
     73             c.a[i+1]+=c.a[i]/10;    
     74             c.a[i]%=10;    
     75         }    
     76     }      
     77     c.len=maxlen+4;      
     78     while (!c.a[c.len]&&c.len>1) c.len--;    
     79     return c;
     80 }
     81 inline gaojing operator - (const gaojing &a,const gaojing &b)//高精-
     82 {
     83     gaojing c;set0(c);
     84     gaojing d;d=a;
     85     for (int i=1;i<=b.len;i++)  
     86       {  
     87         c.a[i]=d.a[i]-b.a[i];  
     88         if (c.a[i]<0)  
     89         {  
     90             c.a[i]+=10;  
     91             int now=i+1;  
     92             while (!d.a[now])  
     93             {  
     94                 d.a[now]=9;  
     95                 now++;  
     96             }  
     97             d.a[now]--;  
     98         }  
     99       }
    100     for (int i=b.len+1;i<=d.len;i++)c.a[i]=d.a[i];  
    101     c.len=d.len;  
    102     while (c.a[c.len]==0&&c.len>1)c.len--;
    103     return c;
    104 }  
    105 inline gaojing operator * (const gaojing &a,const gaojing &b)//高精*
    106 {
    107     gaojing c;set0(c);
    108     for(int i=1;i<=a.len;i++)    
    109       for (int j=1;j<=b.len;j++)    
    110         c.a[i+j-1]+=a.a[i]*b.a[j];    
    111     c.len=a.len+b.len+5;  
    112     for (int i=1;i<=c.len;i++)    
    113       {    
    114         c.a[i+1]+=c.a[i]/10;    
    115         c.a[i]%=10;    
    116       }    
    117     while (!c.a[c.len]&&c.len>1)c.len--;
    118     return c;
    119 }
    120 inline void div_by_2(gaojing &a)
    121 {
    122     for (int i=a.len;i>=1;i--)
    123     {
    124         if (a.a[i]&1 && i!=1)a.a[i-1]+=10;
    125         a.a[i]/=2;
    126     }
    127     while (!a.a[a.len]&&a.len>1)a.len--;
    128 }
    129 inline gaojing operator / (gaojing a,const gaojing &b)//高精/
    130 {
    131     gaojing l,r,ans;
    132     set0(l);l.len=1;
    133     set0(r);r=a;
    134     set0(ans);ans.len=1;
    135     while (l<r||l==r)
    136     {
    137         gaojing mid=l+r;
    138         div_by_2(mid);
    139         if(mid*b==a)return mid;
    140         if(mid*b<a){ans=mid;l=mid+one;}
    141         if(a<mid*b)r=mid-one;
    142     }
    143     return ans;
    144 }
    145 inline gaojing operator ^(const gaojing &a,int p)//高精乘方
    146 {
    147     gaojing ans=one,mult=a;
    148     while (p)
    149     {
    150         if (p&1)ans=ans*mult;
    151         mult=mult*mult;
    152         p>>=1;
    153     }
    154     return ans;
    155 }
    156 inline void chushihua()//初始化,对0、1高精度常数赋值
    157 {
    158     set0(zero); zero.len=1;
    159     set0(one);one.len=1;one.a[1]=1;
    160 }
    161 int main()
    162 {
    163     chushihua();
    164 }
    高精度

     2、打cf必备的一大串头文件(v2.1) ——连快速输出都有了还有什么好怕的

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<algorithm>
     6 #include<cmath>
     7 #include<queue>
     8 #include<deque>
     9 #include<set>
    10 #include<map>
    11 #include<ctime>
    12 #define LL long long
    13 #define inf 0x7ffffff
    14 #define pa pair<int,int>
    15 #define pi 3.1415926535897932384626433832795028841971
    16 using namespace std;
    17 inline LL read()
    18 {
    19     LL x=0,f=1;char ch=getchar();
    20     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    21     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    22     return x*f;
    23 }
    24 inline void write(LL a)
    25 {
    26     if (a<0){printf("-");a=-a;}
    27     if (a>=10)write(a/10);
    28     putchar(a%10+'0');
    29 }
    30 inline void writeln(LL a){write(a);printf("
    ");}
    31 int main()
    32 {
    33 }
    头文件

    3、最短路模板(v1.0) spfa(+slf优化)和dijkstra(+stl堆优化)

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 #include<algorithm>
     5 #define N 100010
     6 #define M 200010
     7 using namespace std;
     8 int n,m,s,t,cnt;
     9 struct edge{int to,next,v;}e[M];
    10 long long dist[N];
    11 bool mrk[N];
    12 int q[N],head[N];
    13 inline void insert(int u,int v,int w)
    14 {
    15     e[++cnt].to=v;
    16     e[cnt].v=w;
    17     e[cnt].next=head[u];
    18     head[u]=cnt;
    19 }
    20 inline void spfa()
    21 {
    22     for (int i=0;i<=n;i++)dist[i]=(long long)1<<40;
    23     q[1]=s;dist[s]=0;mrk[s]=1;
    24     int t=0,w=1;
    25     while (t!=w)
    26     {
    27         if (t==N-1)t=0;else t++;
    28         int now=q[t];
    29         for (int i=head[now];i;i=e[i].next)
    30             if (dist[e[i].to]>dist[now]+e[i].v)
    31             {
    32                 dist[e[i].to]=dist[now]+e[i].v;
    33                 if (!mrk[e[i].to])
    34                 {
    35                     mrk[e[i].to]=1;
    36                     if (dist[e[i].to]<=dist[now])
    37                     {
    38                         q[t]=e[i].to;
    39                         if(t==0)t=N-1;else t--;
    40                         continue;
    41                     }
    42                     if (w==N-1)w=0;else w++;
    43                     q[w]=e[i].to;
    44                 }
    45             }
    46         mrk[now]=0;
    47     }
    48 }
    49 int main()
    50 {
    51     scanf("%d%d%d%d",&n,&m,&s,&t);
    52     for(int i=1;i<=m;i++)
    53     {
    54         int x,y,z;
    55         scanf("%d%d%d",&x,&y,&z);
    56         insert(x,y,z);
    57     }
    58     spfa();
    59     printf("%lld
    ",dist[t]);
    60 }
    spfa
    dijkstra

    ……恩先这样吧还有的再加

    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4098385.html
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