寒假集训_专题四题解_A - k-rounding

题目

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

给出两个数n，k，求最小的这样一个数：这个数是n的倍数，且这个数末尾至少有k个0

题解

分析易得末尾的0由因子2、5组成

代码

#include <iostream>

using namespace std;

int main()
{
	long long a,b;
	cin >> a >> b;
	int x1=0,x2=0;
	long long ans = a;
	while(x1 < b &&ans%2 == 0)
	{
		x1 ++;
		ans >>= 1;
	}
	while(x2 < b && ans % 5 == 0)
	{
		x2 ++;
		ans /= 5;
	}
	x1 = b, x2 = b;
	while(x1--)ans<<=1;
	while(x2--)ans*=5;
	cout << ans;
	return 0;
 }