lcm

Project Euler #5(Smallest positive number divisible by all numbers from 1 to 20): Ways to Optimize? ~Java

由 百般思念 提交于 2021-02-05 20:38:08
问题 Problem 5: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? I have solved the problem 5 of Project Euler Here is the Java code: static long FindLcm(long a,long b) { long lcm,hcf = 0; long i=1; long ger=a>b?a:b; while(i<ger) { if((a%i==0) && (b%i==0)) hcf=i; i++; } lcm=(a*b)/hcf; return lcm; } static void FindMultiple() { long lcm=1; for

Incorrect value of the variable ans that stores the LCM of two numbers (8086 Program)

由 社会主义新天地 提交于 2021-01-05 12:47:21
问题 Following is the code I wrote to find LCM of two numbers in EMU8086. When I ran it, I am getting value 0 in the Ans variable. .MODEL SMALL .DATA Num1 DW 250 Num2 DW 100 Ans DW ? .CODE MOV AX,@DATA MOV DS, AX MOV AX, Num1 MOV BX, Num2 MOV DX, 0000h NEXT: PUSH AX PUSH DX DIV BX CMP DX, 0000h JZ LAST POP DX POP AX ADD AX, Num1 JNC NEXT INC DX JMP NEXT LAST: POP Ans+2 POP Ans MOV AH, 4Ch INT 21h END 回答1: LCM(a, b) = a * b / GCD(a, b) Due to this equation, you can find GCD using Euclid's algorithm

Incorrect value of the variable ans that stores the LCM of two numbers (8086 Program)

由 柔情痞子 提交于 2021-01-05 12:45:03
问题 Following is the code I wrote to find LCM of two numbers in EMU8086. When I ran it, I am getting value 0 in the Ans variable. .MODEL SMALL .DATA Num1 DW 250 Num2 DW 100 Ans DW ? .CODE MOV AX,@DATA MOV DS, AX MOV AX, Num1 MOV BX, Num2 MOV DX, 0000h NEXT: PUSH AX PUSH DX DIV BX CMP DX, 0000h JZ LAST POP DX POP AX ADD AX, Num1 JNC NEXT INC DX JMP NEXT LAST: POP Ans+2 POP Ans MOV AH, 4Ch INT 21h END 回答1: LCM(a, b) = a * b / GCD(a, b) Due to this equation, you can find GCD using Euclid's algorithm

Least Common Multiple

由 喜欢而已 提交于 2020-06-08 03:23:00
问题 I have the current coding which used to be a goto but I was told to not use goto anymore as it is frowned upon. I am having troubles changing it into for say a while loop. I am fairly new to C# and programming in general so some of this is completely new stuff to me. Any help would be appreciated. The actual question is input two numbers and find the lowest common multiple. Here is the original with goto: BOB: if (b < d) { a++; myInt = myInt * a; b = myInt; myInt = myInt / a; if (b % myInt2 =

hrbust 1328

由 丶灬走出姿态 提交于 2020-01-08 13:19:02
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 题目链接： hrbust 1328 这是一道数论的题目，求解方法还是挺巧妙的，尽管够基础。 首先要知道的是多个数的最小公倍数究竟怎么求，有一个公式为lcm(a, b, c) = lcm(lcm(a,b),c)，这个公式对n个数一样也成立。 还有一种求法，是将这多个数均写成素数因子的幂相乘的形式（唯一分解定理），然后对每个素数因子，只取最大的指数，相乘就是最小公倍数了。 这里用的是第二种想法，其实这种想法也是够直观，最接近直觉的。如果A(n)和A(n-1)是相等的，那么说明n的素数因子幂相乘的形式中，没有一个素数因子的幂是大于A(n-1)对应素数的幂的。 而如果n含有两个及以上素数因子，比如(a^x)*(b^y)，那x和y均不会超过A(n-1)对应素数的幂，因为a^x和b^y都是小于n的，A(n-1)中a和b的指数一定大于等于x和y。如果n可以写成a^x这种形式，那么在A(n-1)的a因子的指数一定不会超过x，因为这么大的指数是第一次出现的（特别地，当x为1时，n为素数）。 #include <cstdio> #include <vector> #include <iostream> using namespace std; vector<int> primes; int not_prime[1010]; int

LCM of two numbers

由 混江龙づ霸主 提交于 2020-01-06 07:06:46
问题 I am getting wrong result for my LCM program. Ifirst find gcd of the numbers and then divide the product with gcd. int gcd(int x, int y) { while(y != 0) { int save = y; y = x % y; x = save; } return y; } int lcm(int x, int y) { int prod = x * y; int Gcd = gcd(x,y); int lcm = prod / Gcd; return lcm; } Any help much appreciated. 回答1: Your gcd function will always return 0 . Change return y; to return x; Understand the Euclid's algorithm: RULE 1: gcd(x,0) = x RULE 2: gcd(x,y) = gcd(y,x % y)

LCM of two numbers

由 眉间皱痕 提交于 2020-01-06 07:06:45
问题 I am getting wrong result for my LCM program. Ifirst find gcd of the numbers and then divide the product with gcd. int gcd(int x, int y) { while(y != 0) { int save = y; y = x % y; x = save; } return y; } int lcm(int x, int y) { int prod = x * y; int Gcd = gcd(x,y); int lcm = prod / Gcd; return lcm; } Any help much appreciated. 回答1: Your gcd function will always return 0 . Change return y; to return x; Understand the Euclid's algorithm: RULE 1: gcd(x,0) = x RULE 2: gcd(x,y) = gcd(y,x % y)

Calculate LCM of N numbers modulo 1000000007

由 こ雲淡風輕ζ 提交于 2019-12-29 09:21:10
问题 I was solving following problem on LCM : Calculate LCM of N numbers modulo 1000000007 My approach : typedef unsigned long long ull; const ull mod=1000000007; ull A[10009]; /*Euclidean GCD*/ ull gcd(ull a,ull b) { while( b != 0) { ull t = b; b= a %t; a = t; } return a; } ull lcm(ull a, ull b) { return (a/gcd(a,b))%mod*(b%mod); } ull lcms(int l ,ull * A) { int i; ull result; result = 1; for (i = 0; i < l; i++) result = lcm(result, A[i])%1000000007; return result; } int main() { int T; cin>>T;

How to calculate Least common multiple of {1, 2, 3, …, n}?

由 六月ゝ 毕业季﹏ 提交于 2019-12-21 16:51:26
问题 How to find LCM of {1, 2, ..., n} where 0 < n < 10001 in fastest possible way. The one way is to calculate n! / gcd (1,2,.....,n) but this can be slow as number of testcases are t < 501 and the output should be LCM ( n! ) % 1000000007 Code for the same is: #include<bits/stdc++.h> using namespace std; #define p 1000000007; int fact[10001] = {1}; int gcd[10001] = {1}; int main() { int i, j; for( i = 2;i < 10001; i++){ fact[i] = ( i * fact[i-1]) % p; } for(i=2 ;i < 10001; i++){ gcd[i] =__gcd(

Lowest Common Multiple with doubles in C

由 狂风中的少年 提交于 2019-12-12 18:14:12
问题 I am doing an assignment for a Coursera class that asks me to calculate the Lowest Common Multiple of two numbers, either of which are no larger than 2 * 10 ^ 9. I'm writing this in C and I'm running my code on a test case with the numbers 226553150 and 1023473145. The answer is 46374212988031350, but I'm getting 46374212988031344, which is off by 6! I've written a correct solution in Python that uses essentially the same approach as the one I've posted below, but the details of numeric
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