POJ 1837 balance（dp）

题目描述：
Balance
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 12713 Accepted: 7968
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

题目大意：
天平挂砝码，ci为负值说明在左边，为正在右边，每个挂钩可挂可不挂，但是所有砝码要用上

分析：
这道题我第一反映是dp，就按照这个方向来了
我定义的dp[i][j]:表示挂了前i的砝码之后动力X动力臂+阻力X阻力臂的和
然后最后dp[G][0]就是结果，
和最大为25*20*15=7500，最小为-7500，但是角标不能为负，所以写角标的时候都加上7500，然后结果就是dp[G][7500]

代码：

#include "stdio.h"int C,G;int c[20+5];int w[20+5];int dp[20+1][15000+1];int main(){    scanf("%d%d",&C,&G);    int ci,i,j,k;    for(i=0;i<C;i++)    {        scanf("%d",&c[i]);    }    for(i=1;i<=G;i++)    {        scanf("%d",&w[i]);    }    dp[0][7500]=1;//一个不挂的话也可以平衡    for(i=1;i<=G;i++)    {        for(j=-7500;j<=7500;j++)        {            for(k=0;k<C;k++)            {                if(dp[i-1][j+7500]){//这个if也可以不写，就是算的多点                    dp[i][j+7500+w[i]*c[k]]+=dp[i-1][j+7500];                }            }        }    }    printf("%d\n",dp[G][7500]);    return 0;}