P2651 添加括号III

解题思路：a[1]肯定是分子，a[2]肯定是分母。只要a[1]a[3]…a[n]可以整除a[2]就是可以满足。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double lf;
typedef unsigned long long ull;
typedef pair<ll,int>P;
const int inf = 0x7f7f7f7f;
const ll INF = 1e16;
const int N = 1e6+10;
const ull base = 131;
const ll mod =  1e9+7;
const double PI = acos(-1.0);
const double eps = 1e-4;

inline int read(){
    
    int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
    
    x=x*10+ch-'0';ch=getchar();}return x*f;}
inline string readstring(){
    
    string str;char s=getchar();while(s==' '||s=='\n'||s=='\r'){
    
    s=getchar();}while(s!=' '&&s!='\n'&&s!='\r'){
    
    str+=s;s=getchar();}return str;}
int random(int n){
    
    return (int)(rand()*rand())%n;}
void writestring(string s){
    
    int n = s.size();for(int i = 0;i < n;i++){
    
    printf("%c",s[i]);}}
ll fast_power(ll a,ll p){
    
    
    ll ans = 1;
    while(p){
    
    
        if(p&1) ans = (ans*a)%mod;
        p >>= 1;
        a = (a*a)%mod;
    }
    return ans;
}


int a[N];

void solve(){
    
    
    int n = read();
    for(int i = 1;i <= n;i++){
    
    
        a[i] = read();
    }
    for(int i = 1;i <= n;i++){
    
    
        if(i == 2) continue;
        int x = __gcd(a[2],a[i]);
        a[2] /= x;
    }

    if(a[2] == 1) puts("Yes");
    else puts("No");
}
int main(){
    
    
    srand((unsigned)time(NULL));
    //freopen(  "out.txt","w",stdout);
    int t = read();
    while(t--){
    
    
        solve();
    }
    return 0;
}