Codeforces Round #387(Div. 2)A. Display Size【暴力】

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最新推荐文章于 2024-06-26 12:56:21 发布

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分类专栏：暴力枚举文章标签： Codeforces#387Div. 2

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本文介绍了一个算法问题，即如何寻找最接近正方形的矩形显示布局，使得像素总数等于给定数值n。通过枚举因子的方式，实现了在满足条件的情况下找到行数与列数相差最小的布局方案。

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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

Output

2 4

Input

Output

8 8

Input

Output

1 5

Input

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.

题目大意：

给你一个数n，让你找到两个数i，j，使其乘积为n。

要求输出的是abs（i-j）最小的那个。

思路：

我们一层for枚举i，那么如果i是n的因子，那么对应枚举出j=n/i，然后我们维护最小的abs（i-j）的答案即可。

时间复杂度O（n）；

Ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int ans1,ans2;
        int minn=0x3f3f3f3f;
        for(int i=1;i<=n;i++)
        {
            if(n%i==0)
            {
                int j=n/i;
                if(j>=i)
                {
                    if(j-i<minn)
                    {
                        minn=j-i;
                        ans1=i,ans2=j;
                    }
                }
            }
        }
        printf("%d %d\n",ans1,ans2);
    }
}