LCM = ( num1 * num2 ) / GCD;Source Code: C Program To Find GCD and LCM of Two Numbers#include < stdio.h > int main() { int num1, num2, gcd, lcm, count = 1, small; printf("Enter 2 integer numbers\n"); scanf("%d%d", &num1, &num2); small = (num1 < num2) ?
// C program to find the GCD // (Greatest Common Divisor) of two integers #include <stdio.h> int main() { int num1 = 0; int num2 = 0; int rem = 0; int X = 0; int Y = 0; printf("Enter Number1: "); scanf("%d", &num1); printf("Enter Number2: "); scanf("%d",...
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ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); n = read(); ...
在程序设计中,我们经常会遇到需要求解两个数的最大公约数(Greatest Common Divisor, GCD)和最小公倍数(Least Common Multiple, LCM)的问题。这两个概念在数学上有着重要地位,而在C语言编程中,求解它们同样具有实用价值。接下来,我们将深入探讨如何用C语言计算这两个重要的数学指标。
gcd(b, a % b) : a; } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); n = read(); for(int i = 1; ...
求l c m lcmlcm可以用后缀数组维护一下 。 时间复杂度:O ( n l o g n ) O(nlogn)O(nlogn) AC代码: #include<bits/stdc++.h> using namespace std; const int N=1e5+5; typedef long long ll; ll suf[N],a[N]; ll gcd(ll x,ll y){ ...
Simple C Program to find Normal and Trace of a square matrix in C language with the output and solution.