Time limit per test1 second
memory limit per test 256 megabytes
input standard input
output standard output
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
【翻译】输入n,k，求出最小的数满足能够被n整除且末尾至少有k个0。
题解：
①末尾k个0其实就是能够被10k整除。
②问题转化为求n,k最小公倍数，方法是用LCM(a,b)=a*b/GCD(a,b)
#include<math.h>
#include<stdio.h>
#include<algorithm>
#define ll long long
ll n,k,a,b;
int main()
{
scanf("%I64d%I64d",&n,&k);a=n,b=pow(10,k)+0.5;
printf("%I64d\n",a*b/std::__gcd(a,b));return 0;
}//Paul_Guderian