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  • Codeforces Round #387 (Div. 2) A. Display Size

    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);
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/kuronekonano/p/11794328.html
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