3 条题解
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112209水宇辰 (12209水宇辰) LV 9 @ 2021-12-16 20:39:13
#include<iostream> #include<cmath> using namespace std; int m,n,ans; int gcd(int x,int y) { if(y==0) {return x;} return gcd(y,x%y); } int main() { cin>>n>>m; for(int i=1;i<=sqrt(m*n);i++) { if((n*m)%i==0&&gcd(i,(n*m)/i)==n) ans++; } cout<<ans*2; return 0; }
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12021-12-16 20:31:11@
#include<bits/stdc++.h> using namespace std; int gcd(int a, int b) { return a % b == 0 ? b : gcd(b, a % b);//判断是否互质(最大公约数是否为1) } int main() { int n, m, sum = 0; cin >> n >> m; if (m % n != 0) { cout << 0; return 0; } int temp = m / n; for (int i = 1; i <= floor(sqrt(temp)); i++) { if (temp % i == 0 && gcd(i, temp / i) == 1) sum += 2; } cout << sum; return 0; }
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02021-12-16 20:39:38@
我有两种做法,见下!
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