ural1005 Stone Pile (枚举)

枚举n个石头的状态（即分在左还是右），然后找到答案最小的状态。

#include <bits/stdc++.h>using namespace std;int a[25];int main() {    int n; scanf("%d", &n);    for (int i = 0; i < n; i++) scanf("%d", &a[i]);    int ans = 0x3f3f3f3f;    for (int i = 0; i < (1 << n); i++) {        int l = 0, r = 0;        for (int j = 0; j < n; j++) {            if (i & (1 << j)) l += a[j];            else r += a[j];        }        ans = min(ans, abs(l - r));    }    printf("%d\n", ans);    return 0;}