poj1247 bjfu1239水题

其实就是读题啦，读懂题很简单，就是问一个数组，存不存在一个点，按这个点切成两半，这两半的数字的和是一样的。不多说了，上代码

/*
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
const int MAXN = 40;
int data[MAXN];
int sum1[MAXN], sum2[MAXN];

int main() {
    int n, i;
    while (scanf("%d", &n) == 1 && n > 0) {
        sum1[0] = 0;
        for (i = 1; i <= n; i++) {
            scanf("%d", &data[i]);
            sum1[i] = sum1[i - 1] + data[i];
        }
        sum2[n] = data[n];
        for (i = n - 1; i > 0; i--) {
            sum2[i] = sum2[i + 1] + data[i];
        }
        for (i = 1; i < n; i++) {
            if (sum1[i] == sum2[i + 1]) {
                printf("Sam stops at position %d and Ella stops at position %d.
", i, i + 1);
                break;
            }
        }
        if (i == n) {
            printf("No equal partitioning.
");
        }
    }
    return 0;
}