poj-2707

//408K  0MS G++
#include <cstdio>
#include <cstring>

using namespace std;

int oX;
int oY;

int dX;
int dY;

inline double MIN(double a, double b) {
    return a < b ? a: b;
}

inline double MAX(double a, double b) {
    return a > b ? a: b;
}

void solve(int ox, int oy, int dx, int dy) {
    double case1x = (dx >= ox) ? 1.0 : (double)dx/ox;
    double case1y = (dy >= oy) ? 1.0 : (double)dy/oy;

    double case1 = MIN(case1x, case1y);

    double case2x = (dy >= ox) ? 1.0 : (double)dy/ox;
    double case2y = (dx >= oy) ? 1.0 : (double)dx/oy;

    double case2 = MIN(case2x, case2y);

    printf("%d%%\n", (int)(MAX(case1, case2)*100));

}

int main() {
    while(scanf("%d %d %d %d", &oX, &oY, &dX, &dY) != EOF) {
        if (!oX && !oY && !dX && !dY) {
            return 0;
        }
        solve(oX, oY, dX, dY);
    }
}

水水更健康，就两种放法，找最大的就可以，这道题不能四舍五入。