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P2118 比例简化
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problem/P2118/P2118.cpp
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26
problem/P2118/P2118.cpp
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#include <bits/stdc++.h>
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using namespace std;
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int gcd(int x, int y) {
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if (!y) {
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return x;
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}
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return gcd(y, x % y);
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}
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int main() {
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int a, b, l;
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cin >> a >> b >> l;
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int ansa = l, ansb = 1;
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for (int i = 1; i <= l; i++) {
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for (int j = 1; j <= l; j++) {
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if (gcd(i, j) == 1 && i * b >= j * a && i * ansb < j * ansa) {
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ansa = i;
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ansb = j;
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}
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}
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}
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cout << ansa << ' ' << ansb << endl;
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return 0;
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}
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