SGU - 154 - Factorial （数论）

154. Factorial

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

One number Q written in the input (0<=Q<=10^8).

Output

Write "No solution", if there is no such number N, and N otherwise.

Sample test(s)

Input

2

Output

10

[submit]

[forum]

Author:	Andrew V. Lazarev
Resource:	Saratov Subregional School Team Contest, 2002
Date:	Spring, 2002

思路：对于一个数n的阶乘n！，要求n！的尾部有多少个0........而如果已知尾部有Q个0，则最小的n为多少.......首先在阶乘n！中只要把每个数都拆分为质因数的积，则只需要

质因数5的个数即可，因为2的个数会比5多很多，所以质因数有多少5，尾部就有多少0........

有：

Q = N/5 + N/(5^2) + N/(5^3) + ...

由等比数列求和可得（设只到前k项）：

Q = N(5^k - 1) / [4*(5^k)]，由此得：

N = 4Q * [(5^k)/(5^k-1)]

注意：当Q为0时要输出1

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int fun(int x) {	//求x尾部有多少个0 
	int sum = 0;
	while(x) {
		sum += x / 5;
		x /= 5;
	}
	return sum;
}

int main() {
	int Q;
	while(scanf("%d", &Q) != EOF) {
		if(Q == 0) {
			printf("1\n");
			continue;
		}
		
		int N = Q * 4 / 5 * 5;		//依据公式先大致算到4*Q（除5乘5是为了N正好是5的倍数），然后再往后推 
		while(fun(N) < Q) {
			N += 5;
		}
		
		if(fun(N) == Q) printf("%d\n", N);
		else printf("No solution\n");
	}
	return 0;
}