Codeforces 858A k-rounding

A. k-rounding

time limit per test1 second
memory limit per test256 megabytes

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

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题意：
规定k-rounding数是以k个0结尾的正整数。
给定n和k，求最小的是n的倍数的k-rounding数。

思路：
智障选手从1枚举到了10k，若n的某个倍数%10k为0则说明该数为一个k-rounding数。

代码：

#include<bits/stdc++.h>#define ll long longusing namespace std;int main(){    ll n,k,x;    while(cin>>n>>k)    {        x=1;        while(k--) x*=10;        ll i;        for(i=1;i<=x;i++)        {            if((n*i)%x==0)            {                printf("%lld\n",n*i);                break;            }        }    }    return 0;}

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看了人家的代码后发现自己真的是智障了。
其实就是要求n和10k的最小公倍数而已……QAQ
代码如下

#include<bits/stdc++.h>#define ll long longusing namespace std;int main(){    ll n,k;    while(cin>>n>>k)    {        ll m=1;        while(k--) m*=10;        cout<<n*m/__gcd(n,m)<<endl;    }}