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 first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
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.
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)
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.
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.
Meaning of the questions: Each can be taken from both ends of a number of assumptions made number is x, this is the first time i get, then get the value of i * x, the greatest value is finally ask how much.
. . . Set dp [i] [j] shown on the left to the i, j the right to take a maximum value is obtained, then there is the state transition equation:
dp[i][j] = max(dp[i - 1][j] + val[i] * (i + j), dp[i][j - 1] + val[n - j + 1] * (i + j))
And set the initial state dp [i] [0] (i from 1 to n), dp [0] [j] (j from 1 to n)
#include<cstdio> #include<iostream> #include<algorithm> #define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl #define pii pair<int,int> #define clr(a,b) memset((a),b,sizeof(a)) #define rep(i,a,b) for(int i = a;i < b;i ++) #define pb push_back #define MP make_pair #define LL long long #define ull unsigned LL #define ls i << 1 #define rs (i << 1) + 1 #define INT(t) int t; scanf("%d",&t) using namespace std; const int maxn = 2e3 + 10; int dp[maxn][maxn]; int a[maxn]; int main() { int n; while(~scanf("%d",&n)){ for(int i = 1;i <= n;++ i) scanf("%d",&a[i]); dp[1][0] = a[1]; dp[0][1] = a[n]; rep(i,2,n + 1) dp[i][0] = dp[i - 1][0] + a[i] * i; rep(i,2,n + 1) dp[0][i] = dp[0][i - 1] + a[n - i + 1] * i; for(int i = 1;i <= n;++ i) for(int j = 1;j <= n;++ j) dp[i][j] = max(dp[i - 1][j] + a[i] * (i + j),dp[i][j - 1] + a[n - j + 1] * (i + j)); int ans = 0; for(int i = 0;i <= n;++ i) ans = max(ans,dp[i][n - i]); cout << ans << endl; } return 0; }