Uva 10167

从题目来看，数据量要求不是那么大，对于方法，显然的方法就是简单枚举，经过实际的运算，这样的数据量还是可以的

#include <stdio.h>

#define N 50 * 2 + 1

int x[N], y[N];
int a, b, n;

void split(void);

int main(void)
{
    freopen("in.txt", "r", stdin);
    while(scanf("%d", &n)) {
        if(n == 0) break;
        for(int i = 0; i < 2 * n; i++) {
            scanf("%d%d", &x[i], &y[i]);
        }
        split();
    }
    return 0;
}

int succeed(void)
{
    int up = 0, down = 0;
    int temp;
    for(int i = 0; i < 2 * n; i++) {
        if ((temp = (a * x[i] + b * y[i])) == 0)
            return 0;
        else if (temp > 0)
            up++;
        else down++;
    }
    if(up == down) {
        printf("%d %d\n", a, b);
        return 1;
    }
    else return 0;
}

void split(void)
{
    a = 0; b = 1;
    if (succeed()) {
        return;
    }
    a = 1; b = 0;
    if (succeed()) {
        return;
    }
    for(a = 1; a <= 500; a++) {
        for(b = -500; b <= 500; b++) {
            if (succeed())
                return;
        }
    }
}