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.
題意：給定n個數 每次可以從頭或者尾取出數據 于是按取出來得順序，就可以排成一個數列，假設這個數列為 a1,a2,a3,a4…….an 現在我們假設按照取出來的順序有一個權值 w=a1*1+a2*2+a3*3+….an*n 現在需要編程求出，如何控制取數的順序，讓w的值最大
分析：dp[i][j] 代表從頭取了i個元素 j代表從尾取了多少元素。 注意邊界，即i一個沒取，j一個沒取，別讓i-1<0 或者j-1<0
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int dp[2010][2010];int a[200000];int main(){ int n; while(cin>>n) { for(int i=1;i<=n;i++) { cin>>a[i]; } int ans=0; for(int i=0;i<=n;i++) { for(int j=0;j+i<=n;j++) { if(i>0&&j>0) dp[i][j]=max(dp[i-1][j]+a[i]*(i+j),dp[i][j-1]+a[n+1-j]*(i+j)); else if(i>0) dp[i][j]=dp[i-1][j]+a[i]*i; else if(j>0) dp[i][j]=dp[i][j-1]+a[n-j+1]*j; ans=max(ans,dp[i][j]); } } printf("%d/n",ans ); }}新聞熱點
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