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).

　　Input 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.

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
#define N 105
#define M 1000
using namespace std;int main(){int n,v[N],dp[N][M],a,b,volume,sum;while( scanf("%d",&n)!=EOF  && n>0 ){volume=0;sum=1;for( int i=0;i<n;i++ ){scanf("%d%d",&a,&b);volume+=(a*b);while( b-- )//将单个物品依次加入v[sum++]=a;}sum--;//需要初始化dp中的全部值为0，下一次dp比较的值，上一次可能没有赋值，比较时将与负值（未初始化）进行比较，得到不正确的值//for( int i=0;i<volume;i++ )//    dp[0][i]=0;memset(dp,0,sizeof(dp));for( int i=1;i<=sum;i++ ){for( int j=v[i];j<=volume/2;j++ ){dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+v[i]);//转移方程
            }}printf("%d %d\n",max(dp[sum][volume/2],volume-dp[sum][volume/2]),min(dp[sum][volume/2],volume-dp[sum][volume/2]));}system("pause");return 0;
}

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
#define N 105
#define M 100000
using namespace std;int main(){int n,v[N],dp[M],a,b,volume,sum;while( scanf("%d",&n)!=EOF  && n>0 ){volume=0;sum=1;for( int i=0;i<n;i++ ){scanf("%d%d",&a,&b);volume+=(a*b);while( b-- )//将单个物品依次加入v[sum++]=a;}sum--;//需要初始化dp中的全部值为0，下一次dp比较的值，上一次可能没有赋值，比较时将与负值（未初始化）进行比较，得到不正确的值memset(dp,0,sizeof(dp));for( int i=1;i<=sum;i++ ){for( int j=volume/2;j>=v[i];j-- ){dp[j]=max(dp[j],dp[j-v[i]]+v[i]);//转移方程
            }}printf("%d %d\n",max(dp[volume/2],volume-dp[volume/2]),min(dp[volume/2],volume-dp[volume/2]));}system("pause");return 0;
}