AcWing 282. 石子合并

#include <iostream>
#include <algorithm>
using namespace std;
const int N = ;
int n;
int s[N];//前缀和
int f[N][N];//状态
int main() {
    scanf("%d", &n);
    for (int i = ; i <= n; i ++ ) scanf("%d", &s[i]);
    for (int i = ; i <= n; i ++ ) s[i] += s[i - ];
    //如果区间长度为1，那么合并的代价就为0，所以不用考虑，直接从2开始
    for (int len = ; len <= n; len ++ )//枚举区间长度
        for (int i = ; i + len -  <= n; i ++ ) { //枚举区间起点
            int l = i, r = i + len - ;//左右端点
            f[l][r] = 1e8;//初始化
            for (int k = l; k < r; k ++ )//枚举分界点
                f[l][r] = min(f[l][r], f[l][k] + f[k + ][r] + s[r] - s[l - ]);
        }
    printf("%d\n", f[][n]);
    return ;
}