Manthan, Codefest 16(B--A Trivial Problem)-优快云博客

本文介绍了一道编程题的解决方法，该题要求计算阶乘尾部恰好包含m个零的所有正整数n的个数及具体数值。通过算法分析，找到了规律并给出了实现代码。

Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial ofn ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

input

output

5
5 6 7 8 9

input

output

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that isn! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

题目纠结时间有点长。。

package ManthanCodefest16;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Scanner;

/**
 * Created by lenovo on 2016-03-02.
 */
public class B {
    public static void main(String[] args){
        Scanner scanner = new Scanner(new InputStreamReader(System.in));
        /*
        * 这道题目是数一个数中能被 5 整除的次数，然后每次都能找到这个序列末尾的数，而且除了第一组
        * 其他的都是5个数字一组。恩，就这样
        * */
        int m;
        int[] arr = new int[100000 + 100];
        arr[0] = 4;
        arr[1] = 9;
        int num = 1;
        for(int i = 2; i < arr.length;){
            arr[i] = arr[i-num] + 5;
            num = num(arr[i]+1);
            i += num;
        }
        while(scanner.hasNext()){
            m = scanner.nextInt();
            if(arr[m] == 0){
                System.out.println(0);
            } else {
                if(arr[m] == 4){
                    System.out.println(4);
                    for(int i = 1; i <= arr[m]; ++i){
                        System.out.print(i+" ");
                    }
                } else {
                    System.out.println(5);
                    for(int i = arr[m]-5; i <= arr[m]; ++i){
                        System.out.println(i+" ");
                    }
                }
            }
        }
    }

    public static int num(int n){
        int cnt = 0;
        while(n % 5 == 0 && n != 0){
            n /= 5;
            cnt++;
        }
        return cnt;
    };
}

转载于:https://www.cnblogs.com/ya-cpp/p/5261076.html