Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 5801 | Accepted: 3003 |
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1 //2017-04-06 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 7 using namespace std; 8 9 const int N = 2005; 10 int v[N], n, dp[N][N];//dp[l][r]表示区间l~r间的最大收益 11 //状态转移方程:dp[l][r] = max(dp[l+1][r]+day*v[l], dp[l][r-1]+day*v[r]) 12 13 int dfs(int l, int r, int day) 14 { 15 if(l > r)return 0; 16 if(dp[l][r])return dp[l][r]; 17 if(l == r)return dp[l][r] = day*v[l]; 18 return dp[l][r] = max(dfs(l+1, r, day+1)+day*v[l], dfs(l, r-1, day+1)+day*v[r]); 19 } 20 21 int main() 22 { 23 while(scanf("%d", &n)!=EOF) 24 { 25 for(int i = 0; i < n; i++) 26 scanf("%d", &v[i]); 27 memset(dp, 0, sizeof(dp)); 28 int ans = dfs(0, n-1, 1); 29 printf("%d\n", ans); 30 } 31 32 return 0; 33 }
原文:http://www.cnblogs.com/Penn000/p/6674530.html