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
题意：给出的一系列的数字，可以看成一个双向队列，每次只能从队首或者队尾出队，第n个出队就拿这个数乘以n，最后将和加起来，求最大和 思路：此题不能用贪心做； 反例： 101 1 102 100 由里向外逆推区间，从两端点大小来进行DP； 状态转移方程：dp[i][j]=max(dp[i+1][j]+a[i]+sum[j]-sum[i],dp[i][j-1]+a[j]+sum[j-1]-sum[i-1]); //sum[i]记录的是前i个给出值得和； 代码：
这里写代码#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int a[2005],sum[2005]; int dp[2005][2005]; int main() { int n; while(~scanf("%d",&n)) { memset(sum,0,sizeof(sum)); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; dp[i][i]=a[i]; } for(int k=1;k<n;k++) for(int i=1,j=i+k;j<=n;j++,i++) { dp[i][j]=max(dp[i+1][j]+a[i]+sum[j]-sum[i],dp[i][j-1]+a[j]+sum[j-1]-sum[i-1]); } printf("%d\n",dp[1][n]); } }