public class gcd_lcm { System.out.println(gcd(a, b)); System.out.println(a / gcd(a,b) * b); // LCM = ab / gcd(a,b) } static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } } 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
1266: gcd和lcm(Java) WUSTOJ 1266: gcd和lcm 参考 1naive1的博客 Description 已知a,b的最大公约数为x,也即gcd(a,b)=x; a,b的最小公倍数为y,也即lcm(a,b)=y.给出x,y.求满足要求的a和b一共有多少种。 Input 多组测试样例。每组给两个整数x,y.(1<=x<=100000,1<=y<=100...
*/ 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...
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) = G and lcm(x, y, z) = L? Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least c...
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algorithm algorithms cpp sum coursera data-structures selection-sort gcd longest-common-subsequence fibonacci-numbers binary-search knapsack-problem lcm advanced-data-structures algorithmic-toolbox polynomial-multiplication san-diego advanced-algorithms big-o-notation Updated May 30, 2022 C++ alecrim...
,intb _EuclideanRingElement__gcd(_EuclideanRingElement__m,_EuclideanRingElement__n){while(__n!=0){_EuclideanRingElement__t=__m%__n;__m=__n;__n=__t;}return__m;} I test two c++ program: #include<stdio.h>#include<stdlib.h>typedefunsignedlonglongull;ull nzd(ull a,ull b){if(...
GCD and LCM Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2024 Accepted Submission(s): 904 Problem Description Given two positive integers G and L, could you tell me how many solutions of (最...
The source code to find the GCD is given below. The given program is compiled and executed successfully.// Java program to find the // Greatest Common Divisor import java.util.Scanner; public class Main { public static void main(String[] args) { int num1 = 0; int num2 = 0; int ...
Learn how to find the GCD and LCM of N numbers using C++. This guide provides a step-by-step approach with code examples.