求算法

[chenbing3313]

题目如下：
Suppose you play the following game with your opponent:
There is a sequence of n coins with values v[1], v[2], …, v[n] in a queue which are known in advance. Now each player takes alternate turn to pick a coin. The rule is you can only take the coin from the head or from the tail of the queue each time.
(a) Please design a dynamic program algorithm such that the total value of the coins you pick is the largest. We assume the opponent is very stupid and always makes an incorrect decision. We assume you make the first pick. Pseudo code is not required but you need to show the inductive formula and initial conditions for the program. What is the complexity of your algorithm?
(b) Apply your algorithm to the following sequence: 7, 5, 2, 3, 1, 6.

题干：
A、B二人在一序列中取数。
取数规则：
1、为只能在序列两端取数据。
2、A先取。
3、A、B交替取数据。
3、B只可以去他当前可获取的数据中最小的一个。
目的：给一个算法，使A取得的数据总和最大。

[chenbing3313]

没有答案啊

[dpdp_2012]

细节不知道有没有错，思路肯定没问题，动态规划的思想

#include <stdio.h>

int main()
{
	int n,a[101],p[101][101];
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<=n;i++)
			p[i][0]=0;
		for(int i=1;i<=n;i++)
			p[i][1]=a[i];
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j+i-1<=n;j++)
			{
				int k1,k2;
				if(a[j+1]<a[j+i-1])
					k1=p[j+2][i-2]+a[j];
				else
					k1=p[j+1][i-2]+a[j];
				if(a[j]<a[j+i-2])
					k2=p[j+1][i-2]+a[j+i-1];
				else
					k2=p[j][i-2]+a[j+i-1];
				p[j][i]=k1>k2?k1:k2;
			}
		}
		printf("%d\n",p[1][n]);
	}
}