HDU 4283

题解地址：HDU4283 题解

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
const int maxn =105;
const int INF =0x4f4f4f4f;
int T, kase=0;
int n;
int a[maxn];
int sum[maxn];
int dp[maxn][maxn];

void input(){
    cin>>n;
    sum[0]=0;
    for(int i=1; i<=n; ++i){
        cin>>a[i];
        sum[i]=a[i]+sum[i-1];
    }
    memset(dp, 0 ,sizeof(dp));
    for(int i=1; i<=n; ++i)
    for(int j=i+1; j<=n; ++j)
        dp[i][j]=INF;
}

int main(){
    cin>>T;
    while(T--){
        input();
        for(int len=1;len<n;++len)
            for(int i=1; i+len<=n;++i){
                int j=i+len;
                for(int k = 1; k<=j-i+1; k++)
                    dp[i][j] = min(dp[i][j],
                                   dp[i+1][i+k-1]
                                   +dp[i+k][j]+k*(sum[j]-sum[i+k-1])
                                   +a[i]*(k-1));
            }
        printf("Case #%d: %d\n", ++kase, dp[1][n]);
    }
    return 0;
}