Luogu P2651 add parentheses III

Luogu P2651 add parentheses III

Resolve

• $ A_1 $ is certainly molecule, $ a_2 $ definitely denominator, so as much as possible is $ a_3 $ subsequent changes to the molecule
• $ A_1 / (a_2 / a_3 / a_4 / ...) = a_1 a_3 a_4 ... / a_2 $, so we just make sure $ a_1 a_3 a_4 ... / a_2 $ is a number
• For about minutes, if can be divided into about $ A_2 $ $ $ 1, then it is an integer; each time only $ a_2 = a_2 / gcd (a_2, a_i), i = (1, 3, 4, 5 ... ) $

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
int t,n,s[10005];
int gcd(int a,int b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&s[i]);
        for(int i=3;i<=n;i++) s[2]/=gcd(s[2],s[i]);
        s[2]/=gcd(s[2],s[1]);
        if(s[2]==1) puts("Yes");
        else puts("No");
    }
    return 0;
}