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;
}