B. Multiplication Table
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Sasha grew up and went to first grade. To celebrate this event her mother bought her a multiplication table MM with nn rows and nn columns such that Mij=ai⋅ajMij=ai⋅aj where a1,…,ana1,…,an is some sequence of positive integers.
Of course, the girl decided to take it to school with her. But while she was having lunch, hooligan Grisha erased numbers on the main diagonal and threw away the array a1,…,ana1,…,an. Help Sasha restore the array!
Input
The first line contains a single integer nn (3⩽n⩽1033⩽n⩽103), the size of the table.
The next nn lines contain nn integers each. The jj-th number of the ii-th line contains the number MijMij (1≤Mij≤1091≤Mij≤109). The table has zeroes on the main diagonal, that is, Mii=0Mii=0.
Output
In a single line print nn integers, the original array a1,…,ana1,…,an (1≤ai≤1091≤ai≤109). It is guaranteed that an answer exists. If there are multiple answers, print any.
Examples
input
Copy
5
0 4 6 2 4
4 0 6 2 4
6 6 0 3 6
2 2 3 0 2
4 4 6 2 0
output
Copy
2 2 3 1 2
input
Copy
3
0 99990000 99970002
99990000 0 99980000
99970002 99980000 0
output
Copy
9999 10000 9998
給你一個n*n的矩陣，m[i][j]=a[i]*a[j]，讓你求出a這個數列。
把矩陣用字符表達寫出來，m[i][j]=a[i]*a[j];
就很好發現a[1]=sqrt(m[1][2]*m[1][3]/m[2][3]);
那麼根據m[1][2],m[2][3],...,m[i][i+1],m[n-1][n]就可以求出a[2],a[3],..,a[I+1],a[n].
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#define INF 0x3f3f3f3f
#define LL long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
LL map[1010][1010],a[1010];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin>>map[i][j];
a[1]=sqrt(map[1][2]*map[1][3]/map[2][3]);
cout<<a[1]<<' ';
for(int i=1;i<n;i++)
{
a[i+1]=map[i][i+1]/a[i];
cout<<a[i+1]<<' ';
}
}