题解:

暴力枚举,然后判断是否是矩形

代码:

#include<iostream>
#include<cstdio>
using namespace std;
int n,ans,tot,a[21],dis[21];
int read()
{
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
int main()
{
    n=read();
    for (int i=1;i<=n;i++)a[i]=read();
    for (int i=1;i<=n;i++)
      {
        dis[i+1]=dis[i]+a[i];
        tot+=a[i];
     }
    for (int a=1;a<=n;a++)
     for (int b=a+1;b<=n;b++)
      for (int c=b+1;c<=n;c++)
       for (int d=c+1;d<=n;d++)
        if((dis[b]-dis[a]==dis[d]-dis[c])
        &&(tot-dis[d]+dis[a]==dis[c]-dis[b]))ans++;
    printf("%d",ans);
    return 0;
}

 

posted on 2018-01-02 17:56  宣毅鸣  阅读(95)  评论(0编辑  收藏  举报