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最新推荐文章于 2024-04-01 00:17:56 发布

elijahqi

最新推荐文章于 2024-04-01 00:17:56 发布

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本文介绍了一种名为K-Rounding的算法实现，该算法用于找到一个最小正整数x，使得x以10为基数结束于k或更多个零，并且能够被给定的正整数n整除。通过计算输入整数中因子2和5的数量来确定能够得到的最少零的数量。

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http://www.elijahqi.win/archives/884
A. k-rounding
time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 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
寻找二或五的个数就可以知道十的个数

#include<cstdio>
int n,k,f[10];
int main(){
    freopen("cf.in","r",stdin);
    scanf("%d%d",&n,&k);
    int nn=n;
    while (nn%5==0||nn%2==0){
        if (nn%5==0) nn/=5,f[5]++;
        if (nn%2==0) nn/=2,f[2]++;
    }
    long long tmp=1;
    for (int i=f[5]+1;i<=k;++i) tmp*=5;
    for (int i=f[2]+1;i<=k;++i) tmp*=2;
    tmp*=n;
    printf("%lld",tmp);
    return 0;
}