hiho一下 第173周
阿新 • • 發佈:2017-10-24
turn logs output stdio.h pen 不知道 cte 描述 mov
題目1 : A Game
時間限制:10000ms 單點時限:1000ms 內存限制:256MB描述
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.
輸入
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)
輸出
Output the maximum score Little Ho can get.
- 樣例輸入
-
4 -1 0 100 2
- 樣例輸出
-
99
AC代碼:#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(
WA代碼:
#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; }
思路一樣,就是不知道為什麽不對。
hiho一下 第173周