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:给一个数m表示一个数的阶乘的值后边0的个数,问这样的数有几个并写出来
题解:判断5的个数详解见http://www.cnblogs.com/tonghao/p/4823114.html
因为用二分时区间开的太小没发现 错了几个小时
#include<stdio.h> #include<string.h> #include<string> #include<math.h> #include<algorithm> #define LL long long #define PI atan(1.0)*4 #define DD double #define MAX 100100 #define mod 100 #define dian 1.000000011 #define INF 0x3f3f3f using namespace std; int fun(int x) { int cnt=0; while(x) { cnt+=x/5; x/=5; } return cnt; } int fen(int m) { int left=0; int right=1001000000; int mid,ans; ans=0; while(left<=right) { mid=(left+right)/2; if(fun(mid)>=m) { right=mid-1; //ans=mid; } else left=mid+1; } return left; } int s[MAX]; int main() { int n,m,j,i,t,k; while(scanf("%d",&n)!=EOF) { int sum=0;k=0; int L,R; L=fen(n);R=fen(n+1); // if(L==0||fun(R-1)!=n) // { // printf("0 "); // continue; // } for(i=L;i<R;i++) { if(fun(i)==n) { sum++; s[k++]=i; } } printf("%d ",sum); if(sum!=0) { for(i=0;i<k-1;i++) printf("%d ",s[i]); printf("%d ",s[k-1]); } } return 0; }