比例简化

#include <bits/stdc++.h>
using namespace std;

void solve(){
    double a, b; int L; 
    cin >> a >> b >> L;
    int aa, bb; double eps = 1e6; 
    for(int i = 1; i <= L; i ++){
        for(int j = 1; j <= L; j ++){
            double n = (double)i / j, m = a / b;
            if(n >= m && n - m < eps) aa = i, bb = j, eps = n - m;
        }
    }
    cout << aa << ' ' << bb << endl;
}

int main(){
    solve();
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

void solve(){
    double a, b; int L; 
    cin >> a >> b >> L;
    int aa, bb; double eps = 1e6;
    for(int i = 1; i <= L; i ++){
        int l = 1, r = L;
        double m = a / b;
        while(l < r){
            int mid = l + r + 1 >> 1;
            double n = (double)i / mid;
            if(n >= m) l = mid;  //满足更新左边界
            else r = mid - 1;  //反之更新右边界
        }
        double n = (double)i / l;
        if(n >= m && n - m < eps) aa = i, bb = l, eps = n - m;
    }
    cout << aa << ' ' << bb << endl;
}

int main(){
    solve();
    return 0;
}