Double Factorial——AT

题目描述

For an integer n not less than 0, let us define f(n) as follows:
·f(n)=1 (if n<2)
·f(n)=nf(n−2) (if n≥2)
Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
Constraints
·0≤N≤1018

输入

Input is given from Standard Input in the following format:

N

输出

Print the number of trailing zeros in the decimal notation of f(N).

样例输入 Copy

【样例1】
12
【样例2】
5
【样例3】
1000000000000000000
样例输出 Copy
【样例1】
1
【样例2】
0
【样例3】
124999999999999995

提示

样例1解释
f(12)=12×10×8×6×4×2=46080, which has one trailing zero.
样例2解释
f(5)=5×3×1=15, which has no trailing zeros.

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
ll gcd(ll a,ll b)
{
    ll t;
    while(b!=0)
    {
        t=a%b;
        a=b;
        b=t;
    }
    return a;
}
ll lcm(ll a,ll b)
{
    return a*b/gcd(a,b);
}
start{
    ll n=read;
    if(n%2!=0){
        printf("0\n");
        return 0;
    }
    else{
        ll temp=50;
        ll res=n/10;
        while(temp<=n){
            res+=(n/temp);
            temp*=5;
        }
        cout<<res<<endl;
    }
    end;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:1 ms
    Memory:2024 kb
****************************************************************/