2019 GDUT 新生专题Ⅳ A题

A - k-rounding

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

题目大意是求出一个整数末尾带有大于等于k个0,而且可被n整除。
思路：题目可以转化为求n与10的k次方的最小公倍数。最小公倍数公式=n*m/gcd（n，m）
代码

#include <cstdio>
#include <cstring>//数值有点大，开long long 
int a[10]={0,10,100,1000,10000,100000,1000000,10000000,100000000};
int gcd(int n,int m){
	return m?gcd(m,n%m):n;
}
long long lcm(int n,int m){
	return (long long)n*m/gcd(n,m);
}
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	if(k==0) printf("%d",n);
	else{
		long long res=lcm(n,a[k]);
		printf("%lld",res);
	}
	return 0;
}