POJ 1247 数组首尾求和判定相等 枚举法

    此题把题意弄懂后就很容易，S顺时针走，E逆时针走，确定一个分割点该处两人相遇时途经的客人数之和相等，直接枚举。

#include <iostream>using namespace std;//s顺时针走，e逆时针走，问他们在哪里相遇时路径上面经过的数字之和相等//直接用枚举法int main(){int table[32];int n,sum1,sum2,i,j;while(cin>>n,n!=0){for (i = 1;i <= n;i++)cin>>table[i];for (i = 1;i <= n;i++){sum1 = 0;sum2 = 0;for (j = 1;j <= i;j++) sum1 += table[j];for (j = n;j > i;j--) sum2 += table[j];if (sum1 == sum2){cout<<"Sam stops at position "<<i<<" and Ella stops at position "<<i+1<<"."<<endl;break;}}if (sum1 != sum2){cout<<"No equal partitioning."<<endl;}}return 0;}