36 lines
839 B
C++
36 lines
839 B
C++
|
#include <bits/stdc++.h>
|
|||
|
|
using namespace std;
|
|||
|
|
/*
|
|||
|
|
由于 L 在100以内，因此可以枚举 A′,B′ 的所有组合，然后判断：
|
|||
|
|
(1)、A′,B′ 是否互质；
|
|||
|
|
(2)、A′B′ 是否大于等于 AB，并且最小
|
|||
|
|
|
|||
|
|
时间复杂度：O(L^2 LogL)
|
|||
|
|
*/
|
|||
|
|
int gcd(int a, int b) {
|
|||
|
|
return b ? gcd(b, a % b) : a;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
int main() {
|
|||
|
|
int A, B, L;
|
|||
|
|
cin >> A >> B >> L;
|
|||
|
|
|
|||
|
|
int a, b;
|
|||
|
|
double mi = 1e9;
|
|||
|
|
for (int i = 1; i <= L; i++)
|
|||
|
|
for (int j = 1; j <= L; j++)
|
|||
|
|
if (gcd(i, j) == 1) {
|
|||
|
|
double x = i * 1.0 / j; // A' / B'
|
|||
|
|
double X = A * 1.0 / B; // A/B
|
|||
|
|
|
|||
|
|
if (x >= X && x - X < mi) {
|
|||
|
|
mi = x - X;
|
|||
|
|
a = i, b = j;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
cout << a << ' ' << b << endl;
|
|||
|
|
|
|||
|
|
return 0;
|
|||
|
|
}
|