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RGB7108 - Сондгой тоонуудын үржвэр |
Өгөгдсөн 3 тоон дахь сондгой тоонуудын үржвэрийг ол. Ядаж 1 сондгой тоо байгаа.
Input
Нэг мөрөнд Int төрлийн 3 тоо зайгаар тусгаарлагдан өгөгдөнө.
Output
Үржвэр
Example
Input: 5 2 3 Output: 15
| Нэмсэн: | Bataa |
| Огноо: | 2011-05-24 |
| 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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2021-08-04 11:23:58
hey u stop |
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2021-03-03 07:48:00
#include <stdio.h> int main () { int a,b,c,d,f=1; scanf("%d %d %d",&a,&b,&c ); if(a%2>0) { f=f*a; } if(b%2>0) { f=f*b; } if(c%2>0) { f=f*c; } printf("%d", f); } bat |
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2020-11-08 12:32:06
#include<stdio.h>; main() { int a,b,c; int f=1; scanf("%d %d %d", &a, &b, &c); if((a+1)%2==0) f=f*a ; if((b+1)%2==0) f=f*b; if((c+1)%2==0) f=f*c; printf("%d",f); return 0; } hylbar bodolt |
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2020-11-04 09:05:01
#include<bits/stdc++.h> using namespace std; long long a,d,f,g; int main() { cin>>a>>d>>f; if(a%2>0) a=a; else a=1; if(d%2>0) d=d; else d=1; if(f%2>0) f=f; else f=1; g=a*d*f; cout<<g; return 0; } |
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2020-08-05 09:48:53
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEiM SPAMMING MUAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAHAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Last edit: 2020-08-05 09:50:29 |
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2020-03-30 12:24:17
#include <iostream> using namespace std; int main() { int a, b, c, mulit = 1; cin >> a >> b >> c; if ( a % 2 == 1 ) mulit = mulit * a; if ( b % 2 == 1 ) mulit = mulit * b; if ( c % 2 == 1 ) mulit = mulit * c; cout << mulit <<endl; return 0; } |
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2019-12-25 15:43:29
#include <iostream> using namespace std; int main () { int a,b,c,d; scanf("%d %d %d", &a, &b, &c); if(a%2==0) a=1; else a=a; if(b%2==0) b=1; else b=b; if(c%2==0) c=1; else c=c; d=a*b*c; printf("%d", d); return 0; } \\ingvel hund oilgomjtoi |
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2019-12-13 09:37:42
bi mangar |
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2019-11-19 03:21:30
muu suguud |
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2019-10-22 09:09:48
#include<stdio.h> main(){ int a,b,c,d,f; d=1; scanf("%d%d%d",&a,&b,&c); if(a%2==1){ d=d*a; } if(b%2==1){ d=d*b; } if(c%2==1){ d=d*c; } printf("%d",d); } |
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