B. Multiplication Table （ Codeforces Round #586 (Div. 1 + Div. 2) ）

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define pq priority_queue<int>
#define pql priority_queue<ll>
#define pqn priority_queue<node>
#define v vector<int>
#define vl vector<ll>
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
#define read(x) scanf("%d",&x)
#define lread(x) scanf("%lld",&x);
#define pt(x) printf("%d\n",(x))
#define yes printf("YES\n");
#define no printf("NO\n");
#define gcd __gcd
#define out(x) cout<<x<<endl;
#define over cout<<endl;
#define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
#define input(k) for (int i = 1; i <= (int)(k); i++)  {cin>>a[i] ; }
#define mem(s,t) memset(s,t,sizeof(s))
#define re return ;
#define ok return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod(x) ((x)%9973)
#define test  cout<<"       ok             "<<endl;
#define ls now<<1
#define lli long long
const int mxn = 1e3+5;
lli a[mxn][mxn];
lli dp[mxn];
int gcd(int m, int n)
{
    while(m>0)
    {
        int c = n % m;
        n = m;
        m = c;
    }
    return n;
}
int main()
{
    TLE;
    int n,ans,cnt;
    string str;
    while(cin>>n)
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                cin>>a[i][j];
            }
        dp[1] = sqrt(a[1][2]*a[1][3]/a[2][3] );
        for(int i=2;i<=n;i++)
            dp[i] = a[1][i]/dp[1];
        cout<<dp[1];
        for(int i=2;i<=n;i++)
            cout<<" "<<dp[i];
        cout<<endl;
    }
    ok;
}