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\n";

            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\n");

    else if(ans==d) printf("1\n");

    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\n",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;

}