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  • 喵哈哈村的魔法考试 Round #20 (Div.2) 题解

    题解:

    A 喵哈哈村的跳棋比赛

    题解:其实我们要理解题意就好了,画画图看看这个题意。x<y,那么就交换;x>y,那么x=x%y。

    如果我们经过很多次,或者y<=0了,那么就会无限循环。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    
    long long x,y;
    
    
    void work(){
        while(1){
            if(x==0){
                cout<<y<<endl;
                break;
            }
            if(x<y)swap(x,y);
            if(x==0){
                cout<<y<<endl;
                break;
            }
            if((y<=0)&&(x>y)){
                cout<<"N/A
    ";
                break;
            }
            x%=y;
        }
    }
    
    int main(){
        while(cin>>x>>y)work();
        return 0;
    }
    

    喵哈哈村的扔骰子大赛

    题解:暴力dfs就可以,dfs(i,j),i是当前扔到第j个骰子,j表示当前和的组合,我们把组合用状态压缩表示一下。你本地打表其实也可以

    代码:

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<ctime>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n,m;
    int ans=0;
    inline int Gcd(int a,int b){return b?Gcd(b,(a%b)):a;}
    void go(int v,long long lop){
        if(v>n){
            if(lop&((long long)1<<(long long)m)) ans++;
            return;
        } 
      for(int i=1;i<=6;i++)
        go(v+1,lop|(lop<<i));
    }
    int main(){
        scanf("%d%d",&n,&m);
        go(1,1);
        int d=1;
        for(int i=1;i<=n;i++)d*=6;
        if(ans==0) printf("0
    ");
        else if(ans==d) printf("1
    ");
        else{
            int gc=Gcd(ans,d);
            
            printf("%d/%d",ans/gc,d/gc);
        }
    }
    

    喵哈哈村的魔方大赛

    题解:最多转两次,暴力模拟一下吧

    #include <iostream>
    #include <algorithm>
    using namespace std;
    int F(char f)
    {
        switch(f)
        {
            case 'U':
                return 0;
            case 'D':
                return 1;
            case 'L':
                return 2;
            case 'R':
                return 3;
            case 'F':
                return 4;
            case 'B':
                return 5;
        }
    }
    int k[9][6] = {
        {0, 1, 5, 4, 2, 3}, 
        {0, 1, 3, 2, 5, 4}, 
        {0, 1, 4, 5, 3, 2}, 
        {5, 4, 2, 3, 0, 1}, 
        {1, 0, 2, 3, 5, 4}, 
        {4, 5, 2, 3, 1, 0}, 
        {3, 2, 0, 1, 4, 5}, 
        {1, 0, 3, 2, 4, 5}, 
        {2, 3, 1, 0, 4, 5}
    };
    int main()
    {
        int n, x1, y1, z1, x2, y2, z2, fa1, fa2;
        char f1, f2;
        cin >> n >> x1 >> y1 >> z1 >> f1 >> x2 >> y2 >> z2 >> f2;
        fa1 = F(f1);
        fa2 = F(f2);
        if (x1 == x2 && y1 == y2 && z1 == z2 && fa1 == fa2)
            cout << 0;
        else if (
            x1 == z2 && y1 == y2 && z1 == n + 1 - x2 && fa1 == k[0][fa2] || 
            x1 == n + 1 - x2 && y1 == y2 && z1 == n + 1 - z2 && fa1 == k[1][fa2] || 
            x1 == n + 1 - z2 && y1 == y2 && z1 == x2 && fa1 == k[2][fa2] || 
            x1 == x2 && y1 == n + 1 - z2 && z1 == y2 && fa1 == k[3][fa2] || 
            x1 == x2 && y1 == n + 1 - y2 && z1 == n + 1 - z2 && fa1 == k[4][fa2] || 
            x1 == x2 && y1 == z2 && z1 == n + 1 - y2 && fa1 == k[5][fa2] || 
            x1 == y2 && y1 == n + 1 - x2 && z1 == z2 && fa1 == k[6][fa2] || 
            x1 == n + 1 - x2 && y1 == n + 1 - y2 && z1 == z2 && fa1 == k[7][fa2] || 
            x1 == n + 1 - y2 && y1 == x2 && z1 == z2 && fa1 == k[8][fa2]
        )
            cout << 1;
        else
        {
            if (x1 > (1 + n) / 2)
                x1 = n + 1 - x1;
            if (x2 > (1 + n) / 2)
                x2 = n + 1 - x2;
            if (y1 > (1 + n) / 2)
                y1 = n + 1 - y1;
            if (y2 > (1 + n) / 2)
                y2 = n + 1 - y2;
            if (z1 > (1 + n) / 2)
                z1 = n + 1 - z1;
            if (z2 > (1 + n) / 2)
                z2 = n + 1 - z2;
            if (x1 > y1)
                swap(x1, y1);
            if (x1 > z1)
                swap(x1, z1);
            if (y1 > z1)
                swap(y1, z1);
            if (x2 > y2)
                swap(x2, y2);
            if (x2 > z2)
                swap(x2, z2);
            if (y2 > z2)
                swap(y2, z2);
            if (x1 == x2 && y1 == y2 && z1 == z2)
                cout << 2;
            else
                cout << -1;
        }
        return 0;
    }
    

    喵哈哈村的种树大赛

    题解:简单的线段树的区间更新,区间求gcd的题。

    代码:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    
    int n,m,M;
    int Tree[1048576<<2],tree[1048576];
    
    
    inline int gcd(int x,int y)
    {
        x=x<0?-x:x;  y=y<0?-y:y;
        for (int r;y;r=x%y,x=y,y=r);
        return x;
    }
    
    
    int main()
    {
        scanf("%d%d",&n,&m);
        for(M=1;M<=n+1;M<<=1);
        while(m--)
        {
            int opt;
            scanf("%d",&opt);
            if(opt==1)
            {
                int l,r,t;
                scanf("%d%d%d",&l,&r,&t);
                for(int x=l;x<=n;x+=x&(-x))tree[x]+=t;
                for(int x=r+1;x<=n;x+=x&(-x))tree[x]-=t;
                Tree[l+=M]+=t;
                for(int x=(l>>1);x;x>>=1)Tree[x]=gcd(Tree[x<<1],Tree[(x<<1)|1]);
                Tree[r+=M+1]-=t;
                for(int x=(r>>1);x;x>>=1)Tree[x]=gcd(Tree[x<<1],Tree[(x<<1)|1]);
            }else
            {
                int l,r;
                scanf("%d%d",&l,&r);
                int ret=0,tmp=0;
                for(int x=l+M,y=r+M+1;x^y^1;x>>=1,y>>=1)
                {
                    if(!(x&1))ret=gcd(ret,Tree[x^1]);
                    if(y&1)ret=gcd(ret,Tree[y^1]);
                }
                for(int x=l;x;x-=x&(-x))tmp+=tree[x];
                ret=gcd(ret,tmp);
                printf("%d
    ",ret);
            }
        }
    }
    

    喵哈哈村的玩手机大赛

    题解:最大流,A-手机-B就好了,但是暴力连图是会挂的,所以我们倍增优化一下连接的正方形就好(x

    代码:

    #include<cstdio>
    inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
    const int N=56010,inf=~0U>>2;
    int n,S,T,h[N],gap[N],maxflow;
    struct edge{int t,f;edge *nxt,*pair;}*g[N],*d[N];
    int r,c,a,b,x1,x2,y1,y2,w;
    int i,j,k,fs[62][62][6],ft[62][62][6],pow[8],log[62];
    inline int min(int a,int b){return a<b?a:b;}
    inline void add(int s,int t,int f){
      edge *p=new(edge);p->t=t;p->f=f;p->nxt=g[s];g[s]=p;
      p=new(edge);p->t=s;p->f=0;p->nxt=g[t];
      g[t]=p;g[s]->pair=g[t];g[t]->pair=g[s];
    }
    int sap(int v,int flow){
      if(v==T)return flow;
      int rec=0;
      for(edge *p=d[v];p;p=p->nxt)if(h[v]==h[p->t]+1&&p->f){
        int ret=sap(p->t,min(flow-rec,p->f));
        p->f-=ret;p->pair->f+=ret;d[v]=p;
        if((rec+=ret)==flow)return flow;
      }
      d[v]=g[v];
      if(!(--gap[h[v]]))h[S]=T;
      gap[++h[v]]++;
      return rec;
    }
    int main(){
      for(pow[0]=i=1;i<8;i++)pow[i]=pow[i-1]<<1;
      for(i=1;i<62;i++)for(j=i;j>1;j>>=1,log[i]++);
      read(r);read(c);read(a);read(b);
      for(i=1;i<=r;i++)for(j=1;j<=c;j++){
        fs[i][j][0]=++n,ft[i][j][0]=++n;
        read(k);
        add(fs[i][j][0],ft[i][j][0],k);
      }
      for(k=1;k<6;k++)for(i=1;i<=r;i++)for(j=1;j<=c;j++)if(i+pow[k]-1<=r&&j+pow[k]-1<=c){
        fs[i][j][k]=++n,ft[i][j][k]=++n;
        add(fs[i][j][k],fs[i][j][k-1],inf),add(ft[i][j][k-1],ft[i][j][k],inf);
        add(fs[i][j][k],fs[i+pow[k-1]][j][k-1],inf),add(ft[i+pow[k-1]][j][k-1],ft[i][j][k],inf);
        add(fs[i][j][k],fs[i][j+pow[k-1]][k-1],inf),add(ft[i][j+pow[k-1]][k-1],ft[i][j][k],inf);
        add(fs[i][j][k],fs[i+pow[k-1]][j+pow[k-1]][k-1],inf),add(ft[i+pow[k-1]][j+pow[k-1]][k-1],ft[i][j][k],inf);
      }
      S=n+a+b+1;T=S+1;
      while(a--){
        read(w),read(x1),read(y1),read(x2),read(y2);
        add(S,i=++n,w);
        k=log[j=x2-x1+1];
        add(i,fs[x1][y1][k],inf);
        add(i,fs[x1][y2-pow[k]+1][k],inf);
        add(i,fs[x2-pow[k]+1][y1][k],inf);
        add(i,fs[x2-pow[k]+1][y2-pow[k]+1][k],inf);
      }
      while(b--){
        read(w),read(x1),read(y1),read(x2),read(y2);
        add(i=++n,T,w);
        k=log[j=x2-x1+1];
        add(ft[x1][y1][k],i,inf);
        add(ft[x1][y2-pow[k]+1][k],i,inf);
        add(ft[x2-pow[k]+1][y1][k],i,inf);
        add(ft[x2-pow[k]+1][y2-pow[k]+1][k],i,inf);
      }
      gap[0]=T;
      for(i=0;i++<T;)d[i]=g[i];
      while(h[S]<T)maxflow+=sap(S,inf);
      printf("%d",maxflow);
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/6862281.html
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