hdu 1085 有num1个 1 ，num2个 2 ，num3个 5 （母函数）

有num1个 1 ，num2个 2 ，num3个 5
问它们不能组成的最小正整数是谁

样例的母函数 （1+X）(1+X2)(1+X5+X10+X15)
展开后 X4的系数为0

Sample Input
1 1 3
0 0 0

Sample Output
4

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


int c1[10010], c2[10010] ;
int w[4] = {0 , 1 , 2 , 5};
int num[5];

int main()
{
    //freopen("in.txt","r",stdin) ;
    int n ;
    while(scanf("%d %d %d", &num[1], &num[2], &num[3])!= EOF )
    {
        if (num[1] == 0 && num[2] == 0 && num[3] == 0)
            break ;
        int Max = num[1]*w[1]+num[2]*w[2]+num[3]*w[3];

        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        int i , j , k ;
        for(i=0; i<=w[1]*num[1]; i+=w[1])
            c1[i] = 1;
        int len = w[1]*num[1];
        for(i=2; i<=3; ++i)
        {
            for(j=0; j<=len; ++j)
                for(k=0; k<=w[i]*num[i]; k+=w[i])
                {
                    c2[k+j] += c1[j];
                }
            len += w[i]*num[i];
            for(j=0; j<=len; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        for(i=0; i<=Max; ++i)
            if(c1[i] == 0)
            {
                printf("%d
", i);
                break;
            }
        if(i == Max+1)
            printf("%d
", i);
    }
    return 0;
}