求给的数的gcd就行
#include<bits/stdc++.h>
using namespace std;
vector<int> l(1010);
int gcd(int a, int b)
{
if (a % b == 0)
return b;
else
return gcd(b, a % b);
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> l[i];
for (int i = 2; i <= n; i++)
l[i] = gcd(l[i - 1], l[i]);
cout << l[n];
return 0;
}