Codeforces858a k Rouding

Posted by yjjr's blog on February 6, 2018

标签:模拟

For a given positive integer n denote its k-rounding as the minimum positiveinteger x, such that x ends with k or more zeros in base 10and 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

 

题意:找到一个最小的数,满足下列条件:末尾至少有k个零且能够被n整除

模拟水题

Code

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define LL long long
using namespace std;

LL n,k,cnt5=0,cnt2=0;
int main()
{
    cin>>n>>k;
	while(n%5==0&&cnt5<k){n/=5;cnt5++;}
	while(n%2==0&&cnt2<k){n/=2;cnt2++;}
	rep(i,1,k)n*=10;
	cout<<n<<endl;
	return 0;
}



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