[UR #3] Fusion Reaction Strength

 

    The next greatest common divisor is gcd divided by its smallest prime factor (if any). It can be found that the predecessor gcd of the required sgcd is a divisor of a1, so the prime factorization of a1 can be done directly.

 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=100005;
unordered_map<ll,ll> mmp;
ll gcd(ll x,ll y){ return y?gcd(y,x%y):x;}
int n, 233, num;
ll a,d[233],now;

void dfs(int x,ll y,ll Min){
	if(x==num){ mmp[y]=Min?y/Min:-1; return;}
	dfs(x+1,y,Min),y*=d[x+1];
	if(!Min) Min=d[x+1];
	for(int u=1;u<=c[x+1];u++,y*=(ll)d[x+1]) dfs(x+1,y,Min);
}

inline void prework(ll x){
	for(int i=2;i*(ll)i<=x;i++) if(!(x%i)){
		d[++num]=i;
		for(;!(x%i);c[num]++,x/=i);
	}
	if(x!=1) d[++num]=x,c[num]=1;
	
	dfs(0,1,0);
}

inline void output(){
	printf("%lld ",mmp[a]);
	for(int i=2;i<=n;i++) scanf("%lld",&now),printf("%lld ",mmp[gcd(a,now)]);
}

int main(){
	scanf("%d%lld",&n,&a);
	prework(a);
	output();
	return 0;
}