poj1837 Balance

Description

Gigel has a strange "balance" and he wants to poise it. Actually, the device is different from any other ordinary balance. 
It orders two arms of negligible weight and each arm's length is 15. Some hooks are attached to these arms and Gigel wants to hang up some weights from his collection of G weights (1 <= G <= 20) knowing that these weights have distinct values in the range 1..25. Gigel may droop any weight of any hook but he is forced to use all the weights. 
Finally, Gigel managed to balance the device using the experience he gained at the National Olympiad in Informatics. Now he would like to know in how many ways the device can be balanced. 

Knowing the repartition of the hooks and the set of the weights write a program that calculates the number of possibilities to balance the device. 
It is guaranteed that will exist at least one solution for each test case at the evaluation. 

Input

The input has the following structure: 
• the first line contains the number C (2 <= C <= 20) and the number G (2 <= G <= 20); 
• the next line contains C integer numbers (these numbers are also distinct and sorted in ascending order) in the range -15..15 representing the repartition of the hooks; each number represents the position relative to the center of the balance on the X axis (when no weights are attached the device is balanced and lined up to the X axis; the absolute value of the distances represents the distance between the hook and the balance center and the sign of the numbers determines the arm of the balance to which the hook is attached: '-' for the left arm and '+' for the right arm); 
• on the next line there are G natural, distinct and sorted in ascending order numbers in the range 1..25 representing the weights' values. 

Output

The output contains the number M representing the number of possibilities to poise the balance.

Sample Input

2 4	
-2 3 
3 4 5 8

Sample Output

2

最近做的都是dp，发现状态表达式在dp中很重要，这里用dp[i][j]表示挂满前i个钩码时平衡度为j的方案数，接下来只要用分组背包的思想就行了，这里把砝码分为一组，枚举砝码所能挂的钩码的位置，先初始化dp[i][j]为0，然后令dp[0][7500]=1(注意：因为平衡度的值可能为负，所以先算出极限值7500，然后令7500为平衡值，那么一开始的-7500就变为0不为负数），通过前一个状态为dp[i][j],转移方程是dp[i][j]+=dp[i-1][j-pos[k]*w[i]]。

#include<stdio.h>
#include<string.h>
int pos[30],w[30],dp[30][15050];
int main()
{
	int n,m,i,j,k;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)scanf("%d",&pos[i]);
		for(i=1;i<=m;i++)scanf("%d",&w[i]);
		memset(dp,0,sizeof(dp));
		dp[0][7500]=1;
		for(i=1;i<=m;i++){
			for(j=15000;j>=0;j--){
				for(k=1;k<=n;k++){
					if(j>=pos[k]*w[i] && dp[i-1][j-pos[k]*w[i]]!=0){
					dp[i][j]+=dp[i-1][j-pos[k]*w[i]];
				    }
				}
			}
		}
		printf("%d\n",dp[m][7500]);
	}
	return 0;
}