Description
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.
The treats are interesting for many reasons:
- The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
- Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
- The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
- Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.
Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input
5
1
3
1
5
2
Sample Output
43
Hint
Explanation of the sample:
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
题意:给你一组序列 只能从左端点 或者 右端点 选取一个数乘上选取它的天数 得到一个最大 权值和
思路: dp[i][j] 表示第i个物品 选取j个左边的物品 的最大权值
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
int a[];
int dp[][];
int main(){
ios::sync_with_stdio(false);
int n;
while(cin>>n){
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++){
cin>>a[i];
// sum[i]+=a[i];
}
dp[][]=a[n];
dp[][]=a[];
for(int i=;i<=n;i++){
for(int j=n;j>=n-i+;j--)
dp[i][]+=(a[j]*(n-j+));
for(int j=;j<=i;j++){
if(dp[i-][j-]+a[j]*i<dp[i-][j]+a[n-(i--j)]*i){
dp[i][j]=dp[i-][j]+a[n-(i--j)]*i;
}else{
dp[i][j]=dp[i-][j-]+a[j]*i;
}
}
}
int ans=-inf;
for(int i=;i<=n;i++)
ans=max(dp[n][i],ans);
cout<<ans<<endl;
}
return ;
}