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RGB7316 - Гурван тооны ХБЕХ |
Өгөгдсөн 3 натурал тооны хамгийн бага ерөнхий хуваагдагчийг (ХБЕХ) ол.
Input
Нэг мөрөнд 3 натурал тоо зайгаар тусгаарлагдан өгөгдөнө. Тоонууд 1200-аас ихгүй.
Output
ХБЕХ
Example
Input:
4 6 18
Output:
36
| Нэмсэн: | Bataa |
| Огноо: | 2013-01-14 |
| Хугацааны хязгаарлалт: | 1s |
| Эх кодын хэмжээний хязгаарлалт: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Програмчлалын хэлүүд: | ADA95 ASM32 BASH BF C NCSHARP CSHARP C++ 4.3.2 CPP C99 CLPS LISP sbcl LISP clisp D ERL FORTRAN HASK ICON ICK JAVA JS-RHINO JULIA LUA NEM NICE OCAML PAS-GPC PAS-FPC PERL PHP PIKE PRLG-swi PYTHON PYPY3 PYTHON3 RUBY SCALA SCM guile ST TCL WHITESPACE |
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2023-11-20 12:38:43
dwhgjhgfuufsgygfagsyguyagsgfhgyysgasg hi guy woolcome to my histrory irs s## lol f#rc# its so bab histoyire |
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2023-11-04 11:23:16
#include <iostream> using namespace std; int main() { int a, b, c, min; cin >> a >> b >> c; min = a; if(min > b) min = b; if(min > c) min = c; for(int i = min ; i <= a * b * c ; i++) { if(i % a == 0 && i % b == 0 && i % c == 0) { cout << i; break; } } } |
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2023-10-20 20:00:25
#include <stdio.h> int fun_1(int a, int b) { while (b != 0) { int temp = b; b = a % b; a = temp; } return a; } int fun_2(int a, int b) { return (a * b) / fun_1(a, b); } int main() { int a, b, c, res; scanf("%d %d %d", &a, &b, &c); int res1 = fun_2(a, b); int res2 = fun_2(res1, c); printf("%d", res2); return 0; } ene shig bodson hun algo |
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2023-09-26 08:53:32
LOL Last edit: 2023-09-26 08:53:56 |
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2023-08-19 22:46:23
hope someone willl publish it >:) Last edit: 2023-08-19 22:47:25 |
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2023-05-10 11:48:04
hi no code here >:) bro i said no code here BRO STOP SCROLLING I SAID STOP ALR HERE IS THE CODE HAHA SIKE L BOZO |
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2023-05-09 03:01:15
mr bombastic bomba fantasta Telmuun.B |
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2023-05-02 08:09:24
mr boombastic |
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2023-05-02 08:09:24
mr boombastic |
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2023-03-14 03:14:59
hugatsaanii hyzgaarlalt hetersen geed avdggue arai gej bodsiimsan, durgui hurchle yrshig huul2 hha #include <stdio.h> int main(){ int num1, num2, num3, max, lcm; num1 <= 1200; num2 <= 1200; num3 <= 1200; scanf("%d %d %d", &num1, &num2, &num3); max = (num1 > num2) ? ((num1 > num3) ? num1 : num3) : ((num2 > num3) ? num2 : num3); while(1) { if(max % num1 == 0 && max % num2 == 0 && max % num3 == 0) { lcm = max; break; } max++; } printf("%d", lcm); return 0; } |
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