1 solutions
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0
C :
#include <stdio.h> int main () {int p,r,n,m,temp; scanf("%d %d",&n,&m); if (n<m) {temp=n; n=m; m=temp; //把大数放在n中, 小数放在m中 } p=n*m; //先将n和m的乘积保存在p中, 以便求最小公倍数时用 while (m!=0) //求n和m的最大公约数 {r=n%m; n=m; m=r; } printf("%d",p/n); // p是原来两个整数的乘积 return 0; }
C++ :
#include<iostream> using namespace std; int main(){ int M,N; cin>>M>>N; int x=0; for(int i=1;i<=9999;i++){ int m=M*i; if(m%N==0){ x=m; break; } } cout<<x<<endl; }
Python :
a, b = map(int, input().split()) for i in range(1,(a * b)+1): if i % a == 0 and i % b == 0: print(i) break;
- 1
Information
- ID
- 10832
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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