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。