hdoj1171

//用母函数来做

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int N, M, V, B, Max;
bool s1[250000];        //由于只要求知道存在与否,不要求数值,可以用bool型
int main()
{
    while (scanf("%d", &N) != EOF && N > -1) {
    memset(s1, 0, sizeof(s1));    //初始化
    s1[0] = 1;
    Max = 0;
    for (int i = 0; i < N; ++i) {
        scanf("%d%d", &V, &M);
        for (int j = 0; j <= M; ++j) {
        for (int k = 0; k <= Max / 2; ++k) {
            s1[k + j * V] |= s1[k];    //与或,相当于'加'
        }
        }
        Max += V * M;    //不断地增加最大数值
    }
    B = Max / 2;
    while (!s1[B])        //只要能形成B,就一定能形成A(A=Max-B)
        B--;
    printf("%d %d\n", Max - B, B);

    }
}