GDUT_寒假训练题解报告_数论专题_个人题解报告——题目：A - k-rounding

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GDUT_寒假训练题解报告_数论专题_个人题解报告——题目：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
这个题求n和10的k次方的最小公倍数，这个乱搞就可以了，没啥思路可以说的
完整代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>

using namespace std;
#define INF 0x3f3f3f3f
#define ULL unsigned long long

char n[100010];
int main()
{
	scanf("%s",n);
	int length=strlen(n);
	int N;
	if(length>1)N=(n[length-2]-'0')*10+n[length-1]-'0';
	else N=n[length-1]-'0';

	N%=4;
	int sum=0;
	int k;
	for(int time=1;time<5;time++)
	{
		k=1;
		for(int time1=0;time1<N;time1++)
		{
			k*=time;
		}
		sum+=k;
	}
	printf("%d\n",sum%5);

	return 0;
}