Here is the source code of the C program to find the LCM of two numbers using GCD. The C program is successfully compiled and run on a Linux system. The program output is also shown below. /* * C program to find the LCM of two numbers using GCD */#include<stdio.h>voidmain(){in...
a % b); } // 求两个数的最小公倍数 int lcm(int a, int b) { return a * b / gcd(a, b); } int main() { int num1, num2; printf("Enter two numbers: "); scanf("%d %d", &num1, &num2); int result = lcm(num1, num2); printf...
public class Euclid { public static void main(String[] args) { // TODO Class to find HCF (GCD) of two ints, using recursion Euclid r = new Euclid(); System.out.println(r.hcf(188, 112)); } public int hcf( 浏览4提问于2014-02-08得票数 0 1回答 将Python代码块转换为一行 、、、 ...
printf("Please input two number :\n"); scanf("%d %d",&m,&n); c=gcd(m,n); //c获取最大公约数 d=lcm(m,n); //d获取最小公倍数 printf("The GCD of %d and %d is : %d !\n", m, n, c); printf("The LCM of %d and %d is : %d !\n", m, n, d); return 0; ...
P0295R0 gcd(), lcm() VS 2017 15.3 17 P0298R3 std::byte VS 2017 15.3 17,J P0302R1 Removing Allocator Support In std::function VS 2017 15.5 17 P0307R2 Making Optional Greater Equal Again VS 2017 15.0 P0393R3 Making Variant Greater Equal VS 2017 15.0 P0403R1 ...
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printf<"Pleaseinputtwonumber:\n">; scanf<"%d%d",&m,&n>; c=gcd<m,n>;//c获取最大公约数 d=lcm<m,n>;//d获取最小公倍数 printf<"TheGCDof%dand%dis:%d!\n",m,n,c>; printf<"TheLCMof%dand%dis:%d!\n",m,n,d>;return0;}intgcd<intx,inty>//最大公约数GreatestCommonDivisor{...
—— int k = Gcd(m, n); printf("%d/%d\n", m / k, n / k); int negative = -11 % 5;int positive = negative + 5;—— 正负余数 强制类型转换。(int)时丢失小数点。 计算时间差:将两个时间转换成以分表示,计算以分表示的时间差,然后将差再转换成小时分的形式。—— int hour = d /...
printf(Please input two number :\n); scanf(%d %d,m,n); c=gcd(m,n); //c获取最大公约数 d=lcm(m,n); //d获取最小公倍数 printf(The GCD of %d and %d is : %d !\n, m, n, c); printf(The LCM of %d and %d is : %d !\n, m, n, d); return 0; } int gcd(int x,...
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