标签：-- std first nta array ros ret and string
Little Hi and Little Ho are playing a game. There is an integer array in front of them. They take turns (Little Ho goes first) to select a number from either the beginning or the end of the array. The number will be added to the selecter‘s score and then be removed from the array.
Given the array what is the maximum score Little Ho can get? Note that Little Hi is smart and he always uses the optimal strategy.
The first line contains an integer N denoting the length of the array. (1 ≤ N ≤ 1000)
The second line contains N integers A1, A2, ... AN, denoting the array. (-1000 ≤ Ai ≤ 1000)
Output the maximum score Little Ho can get.
4
-1 0 100 2
99
AC代码：
#include<stdio.h> #include<string.h> #include<stdlib.h> int a[1005], dp[1005][1005], s[1005]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &a[i]); } memset(dp, 0, sizeof(dp)); s[0] = 0; s[1] = a[1]; for (int i = 2; i <= n; i++) { s[i] = s[i - 1] + a[i]; } for (int i = 1; i <= n; i++) { dp[i][i] = a[i]; dp[i][i + 1] = a[i] > a[i + 1] ? a[i] : a[i + 1]; } for (int i = n; i >= 1; i--) { for (int j = i + 2; j <= n; j++) { dp[i][j] = dp[i + 1][j] > dp[i][j - 1] ? s[j] - s[i - 1] - dp[i][j - 1] : s[j] - s[i - 1] - dp[i + 1][j]; } } printf("%d\n", dp[1][n]); return 0; }
WA代码：
#include<stdio.h> #include<string.h> #include<stdlib.h> int a[1005], dp[1005][1005], s[1005]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &a[i]); } memset(dp, 0, sizeof(dp)); s[0] = 0; s[1] = a[1]; for (int i = 2; i <= n; i++) { s[i] = s[i - 1] + a[i]; } for (int i = 1; i <= n; i++) { dp[i][i] = a[i]; dp[i][i + 1] = a[i] > a[i + 1] ? a[i] : a[i + 1]; } for (int i = 3; i <= n; i++) { for (int j = 1; j + i - 1 <= n; j++) { int tmp = -10000000; //j+1->j+i-1 if (a[j] + (s[j + i - 1] - s[j]) - dp[j + 1][j + i - 1] > tmp) { tmp = a[j] + (s[j + i - 1] - s[j]) - dp[j + 1][j + i - 1]; } //j->j+i-2 if (a[j + i - 1] + (s[j + i - 2] - s[j - 1]) - dp[j][j + i - 2] > tmp) { tmp = a[j + i - 1] + (s[j + i - 2] - s[j - 1]) - dp[j][j + i - 2]; } dp[j][j + i - 1] = tmp; } } printf("%d\n", dp[1][n]); return 0; }
思路一样，就是不知道为什么不对。
标签：-- std first nta array ros ret and string
原文地址：http://www.cnblogs.com/dramstadt/p/7722826.html
