经典的石子合并问题！！！
设f[i][j]为从i到j的最大值
然后我们先枚举区间大小，然后枚举起点终点来更新
f[i][j] = min(f[i][k] + f[k+1][j] + sum(i, j));
最后f[1][n]就是答案！！
#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;const int MAXN = 212;
int f[MAXN][MAXN], a[MAXN], s[MAXN], n;int main()
{int ans = 1e9;scanf("%d", &n);REP(i, 1, n + 1) scanf("%d", &a[i]);REP(r, 1, n){swap(a[r], a[r + 1]);REP(i, 1, n + 1) s[i] = s[i-1] + a[i];swap(a[r], a[r + 1]);REP(d, 2, n + 1)for(int st = 1; st + d - 1 <= n; st++){int i = st, j = st + d - 1;f[i][j] = 1e9;REP(k, i, j)f[i][j] = min(f[i][j], f[i][k] + f[k+1][j] + (s[j] - s[i-1]));}ans = min(ans, f[1][n]);}printf("%d\n", ans);return 0;
}