POJ Balance 1837(01背包)

　　　　　　　　　　　　　　　　　　　　　　　　Balance

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 13585 Accepted: 8518

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

Source

Romania OI 2002

题意：一个天平，给出挂钩的位置，负数表示左边，正数表示右边，然后给出钩码的重量，问把所有钩码都挂在天平上，使平衡有多少方案。

分析：力　＝　钩码重力　＊　挂钩的距离

　　　我们就是要找一个平衡状态，设　j=0　时是平衡，j>0右偏，j<0左偏。

　　　由于有负数，距离的范围1～15,钩码最大重量25，最多有20个，所以设　j=20*25*15=7500为平衡

　　　设dp[i][j],i代表前i个钩码挂上去，j代表是右偏还是右偏或者平衡

　　　dp[i][j-w[i]]=dp[i-1][j]; //这个状态由前一个状态得到

           (前i-1个钩码挂上的方案数，如果再在这个基础上在挂一个，并没有增加方案数)

　　　我们就是要找平衡状态的方案数　dp[i][ j+ w[i]*c[k] ] = ∑（dp[i-1][j]）

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int dp[21][15123];int main(){    int n,m,i,j;    int g[22],w[22];    cin>>n>>m;    for(i=1;i<=n;i++)        cin>>g[i];    for(i=1;i<=m;i++)        cin>>w[i];    memset(dp,0,sizeof(dp));    dp[0][7500]=1;    for(i=1;i<=m;i++)    {        for(j=0;j<=15000;j++)        {            if(dp[i-1][j])//优化            {                for(int k=1;k<=n;k++)                {                    dp[i][j+g[k]*w[i]]+=dp[i-1][j];                }            }        }    }    printf("%d\n",dp[m][7500]);    return 0;}