for (int len = 1; len <= n; len++) {         // 區(qū)間長度
    for (int i = 1; i + len - 1 <= n; i++) { // 枚舉起點
        int j = i + len - 1;                 // 區(qū)間終點
        if (len == 1) {
            dp[i][j] = 初始值
            continue;
        }

        for (int k = i; k < j; k++) {        // 枚舉分割點，構(gòu)造狀態(tài)轉(zhuǎn)移方程
            dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + w[i][j]);
        }
    }
}

#include <iostream>
#include <cstring>

using namespace std;

const int N = 307;

int a[N], s[N];
int f[N][N];

int main() {
    int n;
    cin >> n;

    for (int i = 1; i <= n; i ++) {
        cin >> a[i];
        s[i] += s[i - 1] + a[i];
    }

    memset(f, 0x3f, sizeof f);
    // 區(qū)間 DP 枚舉套路：長度+左端點 
    for (int len = 1; len <= n; len ++) { // len表示[i, j]的元素個數(shù)
        for (int i = 1; i + len - 1 <= n; i ++) {
            int j = i + len - 1; // 自動得到右端點
            if (len == 1) {
                f[i][j] = 0;  // 邊界初始化
                continue;
            }

            for (int k = i; k <= j - 1; k ++) { // 必須滿足k + 1 <= j
                f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j] + s[j] - s[i - 1]);
            }
        }
    }

    cout << f[1][n] << endl;


    return 0;
}