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?
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
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.
1
5
5 6 7 8 9
5
0
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.
题意:给一个m,问哪些数的阶乘的末尾0的个数为m;
思路:末尾0的个数为这些数中有多少个5,2的个数一定大于5,所以只需找5的个数就行;
AC代码:
#include <bits/stdc++.h> using namespace std; int main() { int m; scanf("%d",&m); int num,x,sum=0; for(int i=1;i<=100000;i++) { num=1; x=i; while(1) { if(x%5==0) { num++; x=x/5; } else break; } sum+=num; if(sum==m) { cout<<"5"<<" "; for(int j=5*i;j<5*i+5;j++) { printf("%d ",j); } return 0; } else if(sum>m) { break; } } cout<<"0"; return 0; }