C # -- 2nd week experiment -- Task 5 -- compile a console application -- calculate the minimum public multiple and maximum public approx. Of two integers

Source: Internet
Author: User

/* (Start of program header annotation)
* Copyright and version Declaration of the program
* Copyright (c) 2011, a student from the computer College of Yantai University
* All rights reserved.
* File name: enter two integers to obtain the minimum public multiple and maximum public approx.
* Author: Lei hengxin
* Completion date: January 1, September 08, 2012
* Version No.: V1.0
* Description of tasks and Solutions
* Programming philosophy:
* Input description:
* Problem description:
* Program output:
* End the comment in the program Header
Using System;
Using System. Collections. Generic;
Using System. Linq;
Using System. Text;
Namespace ConsoleApplication_do_while
Class Program
Static void Main (string [] args)
Console. WriteLine ("this is a program that inputs two integers to calculate the least common multiple of the two integers and the maximum common number ");
Console. Write ("Enter the first INTEGER :");
String first = Console. ReadLine ();
Console. Write ("enter the second INTEGER :");
String second = Console. ReadLine ();
Int x = int. Parse (first); // type conversion
Int y = int. Parse (second); // type conversion
Int Greatest_common_divisor = gcd (x, y );
Int Least_common_multiple = lcm (x, y );
Console. WriteLine ("The minimum public multiples of {0} and {1} are: {2} The maximum public approx. Is: {3}", x, y, Least_common_multiple, Greatest_common_divisor );
Console. ReadKey ();
// Calculate the maximum public approx.
Static int gcd (int m, int n)
Int I = 2; // define the cyclic control variable
Int Least_common_multiple = 1; // calculates the maximum common approx.
Int min1 = min (m, n );
While (I <= min1)
While (m % I = 0 & n % I = 0) // calculates the common divisor of the denominator.
M = m/I;
N = n/I;
Min1 = min (m, n );
Least_common_multiple = Least_common_multiple * I;
++ I;
Return Least_common_multiple;
// Calculate the minimum public multiple
Static int lcm (int m, int n)
Int Greatest_common_divisor = gcd (m, n); // calculates the maximum public approx.
Int Least_common_multiple = (m/Greatest_common_divisor) * (n/Greatest_common_divisor) * Greatest_common_divisor; // The minimum public multiple has a certain relationship with the maximum public approx.
Return Least_common_multiple;
// Calculate the maximum and minimum values of two numbers
Static int min (int m, int n)
Int min;
If (m> n)
Min = n;
Min = m;
Return min;

Running result:



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