A. Display Size
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly n pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels a and the number of columns of pixels b, so that:
there are exactly n pixels on the display;
the number of rows does not exceed the number of columns, it means a ≤ b;
the difference b - a is as small as possible.
Input
The first line contains the positive integer n (1 ≤ n ≤ 106) — the number of pixels display should have.
Output
Print two integers — the number of rows and columns on the display.
Examples
input
8
output
2 4
input
64
output
8 8
input
5
output
1 5
input
999999
output
999 1001
Note
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
题意：输入显示屏的像素n，输出这是a×b的分辨率屏幕，使a和b之差尽量小。
题解：就是找数n的因数，且两个因数差值尽量小，那么就从sqrt(n)开始找到的第一个可以被n整除的数就是差距最小的两个因数。
#include<stdio.h>
#include<math.h>
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(i=sqrt(n);i>=1;i--)
{
if(n%i==0)
{
break;
}
}
printf("%d %d
",i,n/i);
}
}