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 <a class="tags" href="/tags/TABLE.html" title=table>table</a>[32]; int n,sum1,sum2,i,j; while(cin>>n,n!=0){ for (i = 1;i <= n;i++) cin>><a class="tags" href="/tags/TABLE.html" title=table>table</a>[i]; for (i = 1;i <= n;i++) { sum1 = 0; sum2 = 0; for (j = 1;j <= i;j++) sum1 += <a class="tags" href="/tags/TABLE.html" title=table>table</a>[j]; for (j = n;j > i;j--) sum2 += <a class="tags" href="/tags/TABLE.html" title=table>table</a>[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; }