poj1837 Balance[2차원 01 가방 방법수-천평평형]

제목의 뜻
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
제목: 이미 알고 있는 분동(모두 하나)과 천평이 좌우로 연결된 위치, 천평을 균형 있게 하는 몇 가지 방법이 있는지 묻는다.
방법: 폭력을 생각하기 쉽다. 그러면 안 된다. 그러면 dp[]가 현재의 무게를 표시하는 것으로 생각하기 쉽다. 사고방식은 가방 문제에서 물품의 번호를 표시하는 차원에 정해져 있다. 그리고 전이 방정식을 쓰지 않는다. ==파렴치하게 문제를 뒤집어 풀어야 한다. 그 차원이 필요하면 현재 물품의 번호를 표시한다. 왜냐하면 전이 방정식은 dp[i][j]+=dp[i-1][j-pos[k]*weight[i]]이기 때문이다.
수조, 변수는 자신을 어지럽히고 QAQ를 마이너스로 처리한다. 7500은 중간값인데 이렇게 크지 않으면 어떡하지==코드 보기

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[30][20009];
int c,g,pos[30],weight[30];
int main()
{
   // freopen("cin.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    while(~scanf("%d%d",&c,&g))
    {
        for(int i=1;i<=c;i++)
        {
            scanf("%d",&pos[i]);
            //cc[i]+=15;
        }
        for(int i=1;i<=g;i++)scanf("%d",&weight[i]);
        memset(dp,0,sizeof(dp));
        dp[0][10000]=1;
        int maxn=20000;
        for(int i=1;i<=g;i++)
            for(int j=0;j<=maxn;j++)
                for(int k=1;k<=c;k++)
                    if(j>=pos[k]*weight[i])
                    {
                        dp[i][j]+=dp[i-1][j-pos[k]*weight[i]];
                      //  printf("i=%d,j=%d,k=%d,dp=%d
",i,j,k,dp[i][j]);
                    }

        printf("%d
",dp[g][10000]);
    }
    return 0;
}