using namespace std;
int a[605];
int sum=0;
int main(){
    freopen("frac.in","r",stdin);
    freopen("frac.out","w",stdout);
    int n;
    cin>>n;
    for(int i=0;i<=n-1;i++){
        cin>>a[i];
    }
    for(int i=0;i<=n-1;i++){
        for(int j=0;j<=n;j++){
            if(__gcd(a[i],a[j])==1&&i!=j){
                sum++;
            }
        }
    }
    cout<<sum/2;
    return 0;
}

来自朝鲜的题解

#include <bits/stdc++.h> using namespace std; int a[605]; int sum=0; int main(){ freopen("frac.in","r",stdin); freopen("frac.out","w",stdout); int n; cin>>n; for(int i=0;i<=n-1;i++){ cin>>a[i]; } for(int i=0;i<=n-1;i++){ for(int j=0;j<=n;j++){ if(__gcd(a[i],a[j])==1&&i!=j){ sum++; } } } cout<<sum/2; return 0; }

先写伪代码：

ans=0
for i=[0,n-1)
    for j=[i+1,n)
        if (gcd(a_i,a_j)==1)
            ans+1

正式代码：

#include<iostream>
using namespace std;
int gcd(int a,int b){
    if (b) return gcd(b,a%b);
    return a;
}
int main(){
    freopen("frac.in","r",stdin);
    freopen("frac.out","w",stdout);
    int n;
    cin>>n;
    int a[n+10];
    int cnt=0;
    for (int i=0;i<n;i++) cin>>a[i];
    for (int i=0;i<n-1;i++){
        for (int j=i+1;j<n;j++){
            if (gcd(a[i],a[j])==1){
                cnt++;
            }
        }
    }
    cout<<cnt;
    return 0;
}