Balance

Balance
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 11947 Accepted: 7464

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
题意：
有一个天平，天平左右两边各有若干个钩子，总共有C个钩子，有G个钩码，求将钩码全部挂到钩子上使天平平衡的方法的总数。
其中可以把天枰看做一个以x轴0点作为平衡点的横轴
动态规划dp[i][j],i代表第i个钩码,j代表平衡状态
将所有的钩码都挂上的极端情况就是[-7500,7500],所以移动平衡点,将7500作为新的平衡点,所以对于dp[i-1][j],dp[i][j+len[k]*w[j]]+=dp[i-1][j],有挂i-1个钩码状态转移到挂i个钩码时对应的状态

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;

const int MAX = 15000;

int dp[25][MAX+10];

int w[25],len[25];

int main()
{
    int C,G;
    while(~scanf("%d %d",&C,&G))
    {
        for(int i=1;i<=C;i++)
        {
            scanf("%d",&len[i]);
        }
        for(int i=1;i<=G;i++)
        {
            scanf("%d",&w[i]);
        }
        memset(dp,0,sizeof(dp));
        dp[0][7500]=1;
        for(int i=1;i<=G;i++)
        {
            for(int j=0;j<=MAX;j++)
            {
                if(dp[i-1][j])
                {
                    for(int k=1;k<=C;k++)
                    {
                        dp[i][j+len[k]*w[i]]+=dp[i-1][j];
                    }
                }
            }
        }
        printf("%d\n",dp[G][7500]);
    }
    return 0;
}