题目链接：http://codeforces.com/contest/1220/problem/B
给你一个n*n的矩阵，m[i][j]=a[i]*a[j]，让你求出a这个数列。
把矩阵用字符表达写出来，map[i][j]=a[i]*a[j];
就很好发现a[1]=sqrt(a[1][2]*a[1][3]/a[2][3]);
那么根据a[1]a[2],a[2][3],…,a[i][i+1],a[n-1][n]就可以求出a[2],a[3],…,a[I+1],a[n].
注意要关流，不然会超时
#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()
{
ios::sync_with_stdio(0);
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] << ' ';
}
}