poj 1837 dp有点类似背包

http://poj.org/problem?id=1837

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

题目：给出力臂和各种砝码，求使得天平平衡的方法，要求砝码必须全部用完。

/**
dp[i][j],表示前i个砝码的位置已经确定后，平行点为j的方案数
首先，dp[0][0]=1;转移方程为 dp[i][j+w[i]*pos[k]]+=dp[i-1][j];前提是(dp[i-1][j]!=0)
若 j<0则天平左倾，j>0则天平右倾。最后dp[m][0],即为答案。
又因为数组的下标是不能出现负数的，因此我们刚开始把平衡点看做10000.即dp[0][10000]=1;
因为即使把所有的砝码全部都加在最左面也不过是：10000-20*25*15=2500仍然是整数。
详见代码
*/

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int dp[25][20000],w[25],pos[25];
int n,m;

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&pos[i]);
        for(int i=1;i<=m;i++)
            scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        dp[0][10000]=1;
        for(int i=1;i<=m;i++)
            for(int j=0;j<=20000;j++)
                if(dp[i-1][j])
                    for(int k=1;k<=n;k++)
                        dp[i][j+w[i]*pos[k]]+=dp[i-1][j];
        printf("%d\n",dp[m][10000]);
    }
    return 0;
}