CodeForces - 858A 寒假练习

最新推荐文章于 2020-01-15 11:47:42 发布

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本文介绍了一种称为K-Rounding的算法实现，该算法用于找到能够被给定整数n整除且至少以k个零结尾的最小正整数。通过求解10^k与n的最小公倍数来实现这一目标。

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For a given positive integer n denote its k-rounding as the minimum positive integerx, 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.

Example

Input

375 4

Output

Input

10000 1

Output

Input

38101 0

Output

Input

123456789 8

Output

12345678900000000

代码如下

#include<iostream>
using namespace std;

long long gcd(long long a,long long b)
{
    while(b>0)
    {
        long long c=a%b;
        a=b;
        b=c;
    }return a;
}
long long n,k,s=1;
int main()
{
    //k-rounding 就是有一个K个后置零的数可以被n整除 的 最小数
    //就是求 10^k，n 的最小公倍数
    scanf("%lld%lld",&n,&k);
    for(long long i=1;i<=k;i++)
    {
        s=s*10;
    }gcd(n,s);
    long long lcm;
    lcm=n*s/gcd(n,s);
    printf("%lld\n",lcm);
    return 0;
}