hdu 1171 有num1个w1 ， num2个w2 ……. （母函数）

输入n,代表学院里面有n种设备，并且在下面输入n行，每一行输入v,m代表设备的价格为v,设备的数量是m.然后要求把这些设备的总价值分摊，尽量平分，使其总价值接近相等，最好是相等

比如样例二
（1+X10）(1+X20+X40)(1+X30)
展开后 从sum的一半开始向下枚举 找到第一有系数的X

Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output
20 10
40 40

 

# 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[250010], c2[250010];
int w[55];
int num[55];
int main()
{
    //freopen("in.txt","r",stdin) ;
    int n;
    while(scanf("%d", &n) && n>0)
    {
        memset(w, 0, sizeof(w));
        memset(num, 0, sizeof(num));
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));

        int sum = 0;
        for(int i = 1 ; i <= n ; ++i)
        {
            scanf("%d %d", &w[i], &num[i]);
            sum += w[i]*num[i];
        }

        
        for(int i=0; i<=w[1]*num[1]; i+=w[1])
            c1[i] = 1;
        int len = w[1]*num[1];
        for(int i=2; i<=n; ++i)
        {
            for(int j=0; j<=len; ++j)
                for(int k=0; k<=w[i]*num[i]; k+=w[i])
                {
                    c2[k+j] += c1[j];
                }
            len += w[i]*num[i];
            for(int j=0; j<=len; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        for(int i= sum/2; i>=0; --i)
            if(c1[i] != 0)
            {
                printf("%d %d\n", sum-i, i);
                break;
            }
    }
    return 0;
}