传送门

题意：

给你一个$n\ast n$的矩阵，每个位置由$a_i\ast a_j$得来，主对角线为$0$，让你求出来$a_i$。
$n\le 1e3$

思路：

由公式$\frac{a_{i,j}\ast a_{i,k}}{a_{j,k}}=a_i^2$，显然可以$O(n)$递推出来。

// Problem: B. Multiplication Table
// Contest: Codeforces - Codeforces Round #586 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1220/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
LL a[N][N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);cin>>n;for(int i=0;i<n;i++) {for(int j=0;j<n;j++) {scanf("%d",&a[i][j]);}}for(int i=0;i<n;i++) {int x=(i+1)%n,y=(i+2)%n;printf("%lld ",(LL)sqrt(a[i][x]*a[i][y]/a[x][y]));}return 0;
}
/**/