母函数 杭电1059

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1059

题目大意：价值为1， 2， 3， 4， 5 ， 6 的大理石以及分别有x1, x2, x3, x4, x5, x6个，问两个人是否能平分这些大理石， 要求价值相等？ 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <vector>
using namespace std;

const int MAX = 120010;
int a[MAX], b[MAX];
int main()
{
    int x[10], i, j, k;
    int t = 0;
    while(~scanf("%d %d %d %d %d %d", &x[1], &x[2], &x[3], &x[4], &x[5], &x[6]))
    {
        t++;
        if(x[1] + x[2] + x[3] + x[4] + x[5] + x[6] == 0)
            break;
        //下面这步取mo是关键，不然会超时
        int sum = x[1] % 30 + (x[2] % 30) * 2 + (x[3] % 30) * 3 + (x[4] % 30) * 4 + (x[5] % 30) * 5 + (x[6] % 30) * 6;
        if(sum % 2 == 1)
        {
            printf("Collection #%d:
", t);
            printf("Can't be divided.

");
            continue;
        }
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for(i = 0; i <= x[1]; i++)
        {
            a[i] = 1;
        }
        for(i = 2; i < 7; i++)
        {
            int s;
            for(j = 0; j <= sum / 2; j++)
            {
                for(s = 0, k = 0; s <= x[i] && k + j <= sum / 2; k += i, s++)
                {
                    b[k + j] += a[j];
                }
            }
            for(j = 0; j <= sum / 2; j++)
            {
                a[j] = b[j];
                b[j] = 0;
            }
        }
        if(a[sum / 2] == 0)
        {
            printf("Collection #%d:
", t);
            printf("Can't be divided.

");
        }
        else
        {
            printf("Collection #%d:
", t);
            printf("Can be divided.

");
        }
    }
    return 0;
}