Codeforces Round #313 (Div. 2) A到E Currency System in Geraldion Gerald is into Art Gerald's Hexagon Equivalent Strings

A题，超级大水题，根据有没有1输出-1和1就行了。我沙茶，把%d写成了%n。

B题，也水，两个矩形的长和宽分别加一下，剩下的两个取大的那个，看看是否框得下。

C题，其实也很简单，题目保证了小三角形是正三角形，一个正三角的面积=l*l*(1/2)*cos(30)，由于只要算三角形个数，把六边形扩成一个大三角，剪掉三个小三角，除一下系数就没了。就变成了平方相减。

D题，根据定义递归。然后注意奇数就行了。我沙茶，没加第一种判断dfs(a+len,len,b+len) && dfs(a,len,b)。

E题，有思路，没写代码，就是个DP，设当前走到r，c且不经过黑点的方案数为dp[r][c]，转移的时候用C(r+c-2,r-1)可以算出从(0,0)点走到(r,c)的方案数，然后减去从之前的黑点走到（r，c）的方案数。

总结，好好读题，仔细敲代码

A题

#define HDU

#ifndef HDU

#include<bits/stdc++.h>

#else

#include<cstdio>

#include<cmath>

#include<vector>

#include<map>

#include<set>

#include<algorithm>

#include<cstring>

#endif // HDU

using namespace std;

//const int maxn = 1009;

//int n;

//int a[maxn];

int main()

{

#ifdef local

    freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

#endif // local

    int t;

    int n;

    while(~scanf("%d",&n)){

        bool flag = 0;

        for(int i = 0; i < n; i++){

            scanf("%d",&t);

            if(t == 1) flag = 1;



        }

        if(flag) printf("-1");

        else printf("1");

    }

    return 0;

}



/*

  // for(int i = 0; i < n; i++){

    //}



  scanf("%d",a+i);

    sort(a,a+n);

    if(a[0] == 1 ){



    }else

*/

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B题

#define HDU

#ifndef HDU

#include<bits/stdc++.h>

#else

#include<cstdio>

#include<cmath>

#include<vector>

#include<map>

#include<set>

#include<algorithm>

//#include<iostream>

#endif // HDU

using namespace std;



//#define local

#define PY { puts("YES"); return 0; }

#define PN { puts("NO"); return 0; }



int main()

{

#ifdef local

    freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

#endif // local

    int a0,b0;

    scanf("%d%d",&a0,&b0);

    int a1,b1,a2,b2;

    scanf("%d%d%d%d",&a1,&b1,&a2,&b2);

    int t = (a1+a2),t2 = max(b1,b2);

    if((t<=a0 && t2<= b0)|| (t<=b0 && t2<=a0) ) PY;

    t = (a1+b2); t2 = max(b1,a2);

    if((t<=a0 && t2<= b0)|| (t<=b0 && t2<=a0) ) PY;

    t = (b1+a2); t2 = max(a1,b2);

    if((t<=a0 && t2<= b0)|| (t<=b0 && t2<=a0) ) PY;

    t = (b1+b2); t2 = max(a1,a2);

    if((t<=a0 && t2<= b0)|| (t<=b0 && t2<=a0) ) PY;

    PN;

    return 0;

}

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C题

#define HDU

#ifndef HDU

#include<bits/stdc++.h>

#else

#include<cstdio>

#include<cmath>

#include<vector>

#include<map>

#include<set>

#include<algorithm>

//#include<iostream>

#endif // HDU

using namespace std;



//#define local



int main()

{

#ifdef local

    freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

#endif // local

    int a,b,c,d,e,f,g;

    scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f,&g);

    int l = a+b+c;

    int ans = l*l - a*a - c*c - e*e;

    printf("%d",ans);

    return 0;

}

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D题

#define HDU

#ifndef HDU

#include<bits/stdc++.h>

#else

#include<cstdio>

#include<cmath>

#include<vector>

#include<map>

#include<set>

#include<algorithm>

#include<cstring>

#endif // HDU

using namespace std;



//#define local



bool dfs(char *a,int len,char *b)

{

    if(len == 1) { return (*a) == (*b); }

    if(memcmp(a,b,len) == 0) return true;

    if(len&1) return false;

    len >>= 1;

    return (dfs(a,len,b+len) && dfs(a+len,len,b))||(dfs(a+len,len,b+len) && dfs(a,len,b));

}



const int maxn = 200000+5;

char a[maxn],b[maxn];



int main()

{

#ifdef local

    freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

#endif // local

    scanf("%s%s",a,b);

    int len = strlen(a);

    printf("%s\n",dfs(a,len,b)?"YES":"NO");

    return 0;

}

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