HDU - 1171 Big Event in HDU-优快云博客

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

InputInput contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.
OutputFor each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
Sample Input

Sample Output

20 10
40 40

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<map>
#include<vector>
#include<set>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
int c1[3511330],v[3511330]; //c1存可以组合出来且不大于最大和的一半的值，v标记是否出现过
int num[110],a[110];
int main()
{
  int n;
 while(~scanf("%d",&n))
 {
     if(n<=0) return 0;
     memset(v,0,sizeof(v));
     v[0]=1;
     c1[0]=0;
     int summ=0;
     for(int i=1;i<=n;i++)
     {
         scanf("%d%d",&a[i],&num[i]);
         summ+=num[i]*a[i];
     }
     int sum=summ/2;
     int l=1,r=1;
    for(int i=1;i<=n;i++)
    {
         l=r;
        int u=a[i]*num[i];
        for(int j=a[i];j<=u;j+=a[i])
        {
            for(int k=0;k<l;k++)
            {
                int kk=c1[k]+j;
                if(kk<=sum&&!v[kk])//如果小于一半而且未出现
                {
                    c1[r++]=kk;
                    v[kk]=1;
                }
                if(v[sum]) break;//如果1半也出现过
            }
            if(v[sum]) break;
        }
        if(v[sum]) break;
    }
    sort(c1,c1+r);
    printf("%d %d\n",summ-c1[r-1],c1[r-1]);
 }

}