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). 

2

10 1

20 1

3

10 1

20 2

30 1

-1

20 10

40 40

多重背包问题，可以将问题转化成01背包。两种思路：将多个相同物品拆分成一个一个价值相同的不同物品；也可以在01背包递推时加一层物品个数的循环。
注意这题的坑点！是以一个负整数作为结束，不要想当然以为是-1。因为这个TLE了好久。。以后要认真读题
ps：negative integer负整数 positive integer正整数

//第一种写法，耗时1248ms

#include<stdio.h>

#include<string.h>

int f[],a[];

int max(int x,int y)

{

    return x>y?x:y;

}

int main()

{

    int n,V,sum,c,x,y,i,j;

    while(scanf("%d",&n)&&n>=){

        memset(f,,sizeof(f));

        memset(a,,sizeof(a));

        sum=;c=;

        for(i=;i<=n;i++){

            scanf("%d%d",&x,&y);

            while(y--){

                a[++c]=x;

                sum+=x;

            }

        }

        V=sum/;

        for(i=;i<=c;i++){

            for(j=V;j>=a[i];j--){

                f[j]=max(f[j],f[j-a[i]]+a[i]);

            }

        }

        printf("%d %d\n",sum-f[V],f[V]);

    }

    return ;

}

//第二种写法，耗时811ms

#include<stdio.h>

#include<string.h>

int f[],a[],b[];

int max(int x,int y)

{

    return x>y?x:y;

}

int main()

{

    int n,V,sum,i,j,k;

    while(scanf("%d",&n)&&n>=){

        memset(f,,sizeof(f));

        memset(a,,sizeof(a));

        memset(b,,sizeof(b));

        sum=;

        for(i=;i<=n;i++){

            scanf("%d%d",&a[i],&b[i]);

            sum+=a[i]*b[i];

        }

        V=sum/;

        for(i=;i<=n;i++){

            for(k=;k<=b[i];k++){

                for(j=V;j>=;j--){

                    if(j-a[i]>=){

                        f[j]=max(f[j],f[j-a[i]]+a[i]);

                    }

                }

            }

        }

        printf("%d %d\n",sum-f[V],f[V]);

    }

    return ;

}