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 of n 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 mzeroes. Then print these k integers in increasing order.
Examples
Input 1 Output 55 6 7 8 9 Input 5 Output 0Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
题目给出后缀0个数，输出n!满足条件的所有n值。
数论题。由于因子里含有偶数，所以非零末尾一定是偶数。产生0的因数一定包含5，所以题目就转化为寻找阶乘因子中含有5的个数。
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<set> #include<algorithm> #define MAX 1005 #define INF 0x3f3f3f3f using namespace std; int a[MAX]; int main() { int n,c,i,j; scanf("%d",&n); c=0; for(i=1;i<=1000000;i++){ //注意这里是枚举阶乘的因子，需要大于后缀0最长的情况 int ii=i; while(ii%5==0&&ii>0){ c++; ii/=5; } if(c==n){ printf("5\n"); printf("%d",i); for(j=i+1;j<=i+4;j++){ printf(" %d",j); } break; } else if(c>n){ printf("0\n"); break; } } return 0; }
转载于:https://www.cnblogs.com/yzm10/p/8724453.html
