public class gcd_lcm { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); System.out.println(gcd(a, b)); //最后乘b 放置 a * b 超过int的最大值 System.out.println(a / gcd(a,b) * b); // LCM =...
*/ Java语言: publicclassTestFour{// 最大公约数方法publicstaticintgcd(inta,intb){return(a % b ==0) ? b : gcd(b, a%b); }// 最小公倍数publicstaticintlcm(inta,intb){returna*b/gcd(a, b); }publicstaticvoidmain(String[] args){intnumber1=6, number2 =8; System.out.println(num...
GCD and LCMTime Limit : 2000/1000ms (Java/Other)Memory Limit : 65535/65535K (Java/Other)Total Submission(s) : 1Accepted Submission(s) : 1Problem Descr
hdu 4497 (gcd和lcm的性质+排列组合) GCD and LCM Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Problem Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) ...
HCF/GCD = 90 在Kotlin中查找两个数字的LCM的程序 (Program to find LCM of two numbers in Kotlin) package com.includehelp.basic import java.util.* //Main Function entry Point of Program fun main(args: Array<String>) { //Input Stream ...
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The output of this program is the same as before. We have two functionscompute_gcd()andcompute_lcm(). We require G.C.D. of the numbers to calculate its L.C.M. So,compute_lcm()calls the functioncompute_gcd()to accomplish this. G.C.D. of two numbers can be calculated efficiently ...
import java.util.Scanner; public class LCM_GCD { public static void lcm(int a, int b){ int max, step, lcm = 0; if(a > b){ max = step = a; } else{ max = step = b; } while(a!= 0) { if(max%a == 0 && max%b == 0) { ...
lcm = m*n/gcd lcm*gcd = m*n 1 2 3 4 5则:public class Main { public static void main(String[] args) { System.out.println(gcd(4, 8)); System.out.println(lcm(5, 5)); } //gcd最大公约数 public static long gcd(long a, long b) { if (b == 0) { return a; } else {...