# You Are the One HDU - 4283

## Description

The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?

### 输入描述:

The first line contains a single integer T, the number of test cases. For each case, the first line is n (0 < n <= 100)
The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)

### 输出描述:

For each test case, output the least summary of unhappiness .

Sample Input
2

5
1
2
3
4
5

5
5
4
3
2
2

Sample Output
Case #1: 20
Case #2: 24

## Probelm solving

dp[i][j]=min(dp[i][j],dp[i+1][i+k-1]+dp[i+k][j]+(k-1)*a[i]+k*(sum[j]-sum[i+k-1]))


• $dp[i+1][i+k-1],dp[i+k][j]$为上面分析的三个区间中两个区间的最小花费
• $k*a[i]$表示的是第$i$个数被取走的花费
• $k*(sum[j]-sum[i+k-1])$表示的是选择了$i$之后剩下的所有数在原来最小的基础上要多花费的部分。

## Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define mp make_pair
const ll maxn = 105;
const ll mod = 1e9+7;
const ll INF = 0x3f3f3f3f;
ll d[4][2]={1,0,0,1,-1,0,0,-1};
ll dp[maxn][maxn],a[maxn],sum[maxn];
int main() {
ios::sync_with_stdio(0);
#ifdef Uncle_drew
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#else
#endif
ll t,n,cas=0;
cin>>t;
while(t--){
cas++;
cin>>n;
memset(dp,0,sizeof(dp));
memset(sum,0,sizeof(sum));
for(ll i=1;i<=n;i++){
for(ll j=i+1;j<=n;j++){
dp[i][j]=INF;
}
}
for(ll i=1;i<=n;i++){
cin>>a[i];
sum[i]=sum[i-1]+a[i];
}
for(ll l=2;l<=n;l++){
for(ll i=1;i+l<=n+1;i++){
ll j=i+l-1;
for(ll 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-1)*a[i]+k*(sum[j]-sum[i+k-1]));
}
}
}
cout<<"Case #"<<cas<<": "<<dp[1][n]<<endl;
}
return 0;
}