思路：環狀的直接在n後面加上a[0]-a[n]變成鏈狀即可。
這題範圍小，如果n<1000,則必須四邊形不等式優化降低複雜度為O(n^2)
Code：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int AX = 4e2+6;
int dp[AX][AX];
int a[AX];
int s[AX][AX];
int sum[AX];
int n ;

int main(){
    scanf("%d",&n);
    sum[0] = 0;
    for( int
 
 i = 1 ; i <= n ; i++ ){
        scanf("%d",&a[i]);
        sum[i] = sum[i-1] + a[i] ;
    }
    for( int i = n + 1 ; i <= 2 * n ; i++ ){
        sum[i] = sum[i-1] + a[i-n];
    }
    for( int i = 1 ; i <= 2 * n ; i++ ){
        dp[i][i] = 0;
        s[i][i] = i;
    }
    for( int r = 2
 
 ; r <= n ; r++ ){
        for( int i = 0 ; i + r - 1 <= 2 * n ; i++ ){
            int j = i + r - 1 ; 
            dp[i][j] = INF;
            int tmp = sum[j] - sum[i-1] ;
            for( int k = s[i][j-1] ; k <= s[i+1][j] ; k++ ){
                //dp[i][j] = min( dp[i][j] , dp[i][k] + dp[k+1][j] + tmp );
 

                if( dp[i][j] > dp[i][k] + dp[k+1][j] + tmp ){
                    dp[i][j] = dp[i][k] + dp[k+1][j] + tmp;
                    s[i][j] = k ;
                }
            }
        }
    }
    int res = INF;
    for( int i = 1 ; i <= n ; i++ ){
        res = min( res , dp[i][i+n-1] ) ;
    }
    printf("%d\n",res);
    return 0 ;
}

應用：
s[i][j]記錄dp[i][j]取得最優解的k，每次只需要將k從s[i][j-1] 到 s[i+1][j]列舉即可。