西南民族大学第十届校赛-解题记录

重现地址：牛客重现赛

A.dreamstart的催促

这是快速幂的模板题，注意$(a + b)\% c = ((a\% c) + (b\% c))\% c$

# include <iostream>
# include <cstdio>
using namespace std;
const int maxn = 1e5+5;
const long long MOD = 10000019;
long long num[maxn];
int N;
void Init()
{
    for(int i=0;i<N;i++)
        scanf("%lld",&num[i]);
}
long long getnum(long long a,long long b)
{
    long long res = 1;
    while(b)
    {
        if(b&1) res = res*a%MOD;
        a = a*a%MOD;
        b >>= 1;
    }
    return res;
}
void Solve()
{
    long long ans = 0;
    for(int i=0;i<N;i++)
        ans += getnum(num[i],i+1);
    ans = ans%MOD;
    printf("%lld
",ans);
}
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        Init();
        Solve();
    }  
    return 0;
}

 B.TRDD got lost again

一看就是BFS的搜索题，唯一特殊的移动的步幅为2。同时注意有两个坑：1.输入包含空格，用scanf会超时；2.题目卡空间，标记数组vis用bool不要用int

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
using namespace std;
const int maxn = 6005;
char Map[maxn][maxn],c;
int N,M;
bool vis[maxn][maxn];
int Sx,Sy,Tx,Ty;
int dx[4] = {0,2,0,-2};
int dy[4] = {2,0,-2,0};
struct Node
{
    int x,y,t;
    Node(){}
    Node(int xx,int yy,int tt)
    {
        x = xx;
        y = yy;
        t = tt;
    }
};
void Init()
{
    N = N*2+1;
    M = M*2+1;
    for(int i=0;i<N;i++)
    {
        getchar();
        for(int j=0;j<M;j++)
        {
            Map[i][j]=getchar();
            if(Map[i][j]=='S')
            {
                Sx=i;
                Sy=j;
            }
            if(Map[i][j]=='T')
            {
                Tx=i;
                Ty=j;
            }
        }
    }
    memset(vis,false,sizeof(vis));
}
void Solve()
{
    int ans = -1;
    queue<Node> Q;
    Q.push(Node(Sx,Sy,1));
    vis[Sx][Sy] = 1;
    while(!Q.empty())
    {
        Node temp = Q.front();
        Q.pop();
        if(Map[temp.x][temp.y] == 'T')
        {
            ans = temp.t;
            break;
        }
        for(int i=0;i<4;i++)
        {
            int nx = temp.x+dx[i];
            int ny = temp.y+dy[i];
            int tempx = temp.x+dx[i]/2;
            int tempy = temp.y+dy[i]/2;
            if(nx>=1&&nx<N&&ny>=1&&ny<M)
            {
                if((Map[tempx][tempy]==' ')&&(Map[nx][ny]==' '||Map[nx][ny]=='T'))
                    if(!vis[nx][ny])
                    {
                        Q.push(Node(nx,ny,temp.t+1));
                        vis[nx][ny] = true;
                    }
            }
        }  
    }
    if(ans>0)
        printf("%d
",ans);
    else
        printf("TRDD Got lost...TAT
");   
}
int main()
{
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        Init();
        Solve();
    }
    return 0;
}

C.Company

所有的村庄是一个树的结构，某个节点的答案来自于两个部分：1.该节点的所有子节点的答案和；2.该节点自身是否劳动力不足，不足就为1，否则为0

# include <iostream>
# include <cstdio>
# include <vector>
# include <cstring>
using namespace std;
const int maxn = 2e5+10;
int num[maxn],vis[maxn],ans[maxn];
vector<int> G[maxn];
int n,k;
void Init()
{
    int u,v;
    for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);
    for(int i=0;i<maxn;i++)
        G[i].clear();
    for(int i=0;i<n-1;i++)
    {
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    memset(vis,0,sizeof(vis));
    memset(ans,0,sizeof(ans));
}
void dfs(int x)
{
    if(num[x]<=k)
        ans[x]++;
    for(int i=0;i<G[x].size();i++)
    {
        if(!vis[G[x][i]])
        {
            vis[G[x][i]] = 1;
            dfs(G[x][i]);
            ans[x] += ans[G[x][i]];    
        }        
    }
}
void Solve()
{
    vis[1] = 1;
    dfs(1);
    for(int i=1;i<=n;i++)
    {
        if(i<n)
            printf("%d ",ans[i]);
        else
            printf("%d
",ans[i]);
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        Init();
        Solve();
    }    
    return 0;
}

D.A-B-C

一个简单模拟题

# include <iostream>
# include <cstdio>
using namespace std;
const int maxn = 5005;
int N,num[maxn];
void Init()
{
    for(int i=1;i<=N;i++)
        scanf("%d",&num[i]);
}
void Solve()
{
    int f = 0;
    for(int i=1;i<=N;i++)
    {
        int l = num[i];
        int ll = num[l];
        if(i==num[ll])
        {
            f = 1;
            break;
        }    
    }
    if(f) printf("YES
");
    else printf("NO
");
}
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        Init();
        Solve();
    }
    return 0;
}

E.PPY的字符串

模拟题，我用结构体存储字符串的信息,112333-》(2,1)(1,2)(3,3)

# include <iostream>
# include <cstdio>
# include <cstring>
using namespace std;
const int maxn = 1e5;
struct Node
{
    char c;
    int n;
    Node(){}
    Node(char cc,int nn)
    {
        c = cc;
        n = nn;
    }
};
Node ss[maxn];
char s[maxn];
int N;
void change()
{
    int k = 0;
    char tempnum = s[0];
    int tempcnt = 1;
    int L = strlen(s);
    for(int i=1;i<L;i++)
    {
        if(s[i]==tempnum)
            tempcnt++;
        else
        {
            ss[k++] = Node(tempnum,tempcnt);
            tempnum = s[i];
            tempcnt = 1;
        } 
    }
    ss[k++] = Node(tempnum,tempcnt);
    int f = 0;
    for(int i=0;i<k;i++)
    {
        s[f++] = char(ss[i].n+'0');
        s[f++] = ss[i].c;
    }
    s[f] = '';
}
void Solve()
{
    for(int i=0;i<N-1;i++)
        change();
    printf("%d %s
",strlen(s),s);
}
int main()
{
    while(scanf("%s%d",s,&N)!=EOF)
    {
        Solve();
    }
    return 0;
}

F.集训队的脱单大法

数据很水，暴力求解

# include <iostream>
# include <cstdio>
# include <cmath>
using namespace std;
const int maxn = 1e5+5;
int num[maxn];
int n;
void Init()
{
    for(int i=0;i<n;i++)
        scanf("%d",&num[i]);
}
int getmax(int l,int r)
{
    int res = -1;
    for(int i=l;i<=r;i++)
        res = max(res,num[i]);
    return res;
}
void Solve()
{
    int res = -1;
    for(int i=0;i<n-1;i++)
    {
        int Lmax = getmax(0,i);
        int Rmax = getmax(i+1,n-1);
        res = max(res,abs(Lmax-Rmax));
    }
    printf("%d
",res);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        Init();
        Solve();
    }    
    return 0;
}

G.不想再WA了

数据很水，直接dfs

# include <iostream>
# include <cstdio>
using namespace std;
const int maxn = 105;
int cnt,N,s[maxn];
void Init()
{
    scanf("%d",&N);
    cnt = 0;    
}
bool judge()
{
    for(int i=0;i+1<N;i++)
        if(s[i]==3&&s[i+1]==1)
            return false;
    return true;
}
void dfs(int len)
{
    if(len==N)
    {
        if(judge())    cnt++;
        return;        
    }
    else
    {
        for(int i=1;i<=3;i++)
        {
            s[len] = i;
            dfs(len+1);
        }    
    }
}
void Solve()
{
    dfs(0);
    printf("%d
",cnt);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        Init();
        Solve();            
    }
    return 0;
}

H.Ricky’s RealDan’s Ricky

博弈题，找规律，只有娃娃机个数为 1 个， 并且里面的娃娃只有偶数个， 第一个人才能赢 

# include <iostream>
# include <cstdio>
using namespace std;
const int maxn = 1e5+5;
int num[maxn];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        if(n==1&&num[0]%2==0)
            printf("Ricky is Winner
");
        else
            printf("RealDan is Winner
");
    }    
    return 0;
}

I.小A的期末作业

纯模拟，注意最后输出50个就要加一个空行

# include <iostream>
# include <cstdio>
# include <cmath>
using namespace std;
int main()
{
    int N;
    while(scanf("%d",&N)!=EOF)
    {
        for(int i=0;i<2*N-1;i++)
        {
            int spL = N-1-abs(N-1-i);
            for(int j=0;j<spL;j++)
                printf(" ");
            for(int j=0;j<N;j++)
                printf("*");
            printf("
");
        }
    }
    return 0;
}

J.怪盗基德 & 月之瞳宝石

模拟题，在星体和能源体的分布上，遍历每一个星体，找到与这个星体最近能源体的距离，在所有的最近距离中取最大的就是答案

# include <iostream>
# include <cstdio>
# include <algorithm> 
# include <cmath>
using namespace std;
const int maxn = 1e5+10;
long long s[maxn],p[maxn]; 
int n,m;
void Init()
{
    for(int i=0;i<n;i++)
        scanf("%lld",&s[i]);
    for(int i=0;i<m;i++)
        scanf("%lld",&p[i]);    
}
void Solve()
{
    sort(s,s+n);
    sort(p,p+m);
    long long ans,temp;
    int j = 0;
    ans = -1;
    for(int i=0;i<n;i++)
    {
        while(p[j]<s[i]&&j<m-1) j++;
        //printf("%d %d %d %d
",j,p[j-1],s[i],p[j]);
        if(j==0) temp = abs(p[j] - s[i]);
        else temp = min(abs(p[j] - s[i]),abs(s[i] - p[j-1]));
        //printf("%lld
",temp);
        ans = max(ans,temp);
    }    
    printf("%lld
",ans);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        Init();
        Solve();
    } 
    return 0;
}

K.正方体

模拟题

# include <iostream>
# include <cstdio>
using namespace std;
int Map[4][5];
void Init()
{
    for(int i=0;i<3;i++)
        for(int j=0;j<4;j++)
            scanf("%d",&Map[i][j]);
}
void Solve()
{
    int a,b;
    for(int i=0;i<4;i++)
    {
        if(Map[0][i]!=0) a = Map[0][i];
        if(Map[2][i]!=0) b = Map[2][i];
    }
    if(a!=b||Map[1][0]!=Map[1][2]||Map[1][1]!=Map[1][3])
        printf("No!
");
    else
        printf("Yes!
");
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int i=0;i<T;i++)
    {
        Init();
        Solve();
        if((i+1)%50==0)
            printf("
");
    }
    return 0;
}

L.简单的分数

运用gcd公式通分，注意处理负号的问题

# include <iostream>
# include <cstdio>
# include <cmath>
using namespace std;
int op,a,b,c,d;
void Init()
{
    scanf("%d%d%d%d%d",&op,&a,&b,&c,&d);    
}
int gcd(int x,int y)
{
    return y?gcd(y,x%y):x;
}
void Solve()
{
    int bd = b*d;
    int adbc = op==1?a*d+b*c:a*d-b*c;
    int temp = gcd(abs(bd),abs(adbc));
    printf("%d/%d
",adbc/temp,bd/temp);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        Init();
        Solve();
    }    
    return 0;
}

M.HJ浇花

运用差分标记能够很快很巧妙的处理区间处理的问题吗，推荐一个讲的清楚的博客

# include <iostream>
# include <cstdio>
# include <cstring>
using namespace std;
const int maxn = 1e6+10;
int num[maxn],sum[maxn],cnt[maxn];
int n;
void Init()
{
    memset(cnt,0,sizeof(cnt));
    memset(sum,0,sizeof(sum));
    memset(num,0,sizeof(num));
}
void Solve()
{
    int l,r;
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&l,&r);
        sum[l]++;
        sum[r+1]--;
    }
    int res = 0;
    num[0] = sum[0];
    cnt[num[0]]++;
    for(int i=1;i<maxn;i++)
    {
        num[i] = num[i-1] + sum[i];
        cnt[num[i]]++;
    }
    for(int i=1;i<n;i++)
        printf("%d ",cnt[i]);
    printf("%d
",cnt[n]);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        Init();
        Solve();
    }
    return 0;
}