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

int main()
{
    int a, b, c;
    cin >> a >> b >> c;
    cout << __gcd(a, __gcd(b, c)) << endl;
    return 0;
}

#include <iostream>
using namespace std;
short a,b,c;
int main()
{
	cin >> a >> b >> c;
	short maxx = 0;
	for (int i = 1;i < 5001;i++)
	{
		if (a % i == 0 && b % i == 0 && c % i == 0)
			maxx = i;
	}
	cout << maxx;
	return 0;
}
//存储答案

#include<bits/stdc++.h> using namespace std; int a,b; int gcd(int a,int b){ if(b==0)return a; return gcd(b,a%b); } int main() { cin>>a>>b; if(a<b)swap(a,b); gcd(a,b); cout<<b; return 0; }

欧几里得解题10秒一题：

#include<iostream>
#define int long long
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
signed main(){
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    int ab=gcd(a,b);
    printf("%d",gcd(ab,c));
    return 0;
}

先求出最小的两个数的公约数，再与另一个数求公约数

#include<iostream>
using namespace std;
int a,b,c,ans;
int gy(int x,int y){
	int r=x%y;
	while(r){
		x=y;
		y=r;
		r=x%y;
	}
	return y;
}
void px(int x,int y,int z){
	x=min(a,min(b,c));
	z=max(a,max(b,c));
	y=a+b+c-x-z;
	a=x;b=y;c=z;
	return;
}
int main(){
	cin>>a>>b>>c;
	px(a,b,c);
	ans=gy(a,b);
	ans=gy(ans,c);
	cout<<ans;
	return 0;
}