hiho一下第173周

题目1 : A Game

时间限制:10000ms

单点时限:1000ms

内存限制: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.

输入

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)

输出

Output the maximum score Little Ho can get.

样例输入

4
-1 0 100 2

样例输出

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_JUDGEfreopen("input.txt", "r", stdin);
#endifint 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_JUDGEfreopen("input.txt", "r", stdin);
#endifint 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-1if (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-2if (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。

几个小时之后，两种都不对了。那你特么倒是解释解释这是怎么回事啊？

真操蛋，这种问题遇到无数次了。

OK了，辣鸡OJ评测机有问题。

转载于:https://www.cnblogs.com/dramstadt/p/7722826.html