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

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

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