Leshan normal programming competition 2020-h: least common multiple [find the number of factors]

Title Description

Given an integer b, in addition a Express 1 To 1018 All integers in , Calculation formula [a, b] / a How many different results , here [a, b] Represents an integer a And integers b The least common multiple of .

Input

The input contains only one set of data ;
The first line contains an integer b （1 ≤ b ≤ 1010）.

Output

Output the number of different results of the formula .

The sample input

2

Sample output

2

Tips

Arbitrary positive integer a And 2 The least common multiple of must be equal to 2 * a perhaps a, So formula [a, b] / a The result can only be 1 perhaps 2.

Ideas

Enumerate more b, We will find that it is to ask b The number of factors .
for example b = 4;
(4, a) == > The least common multiple may be (a,2a,4a), Divided by a for (1,2,4), Observation shows that these are 4 Factor of .
So directly find the number of factors , however b There's a lot of scope , Observe again [1,sqrt(b)] There is this n Half of the factor ,(sqrt(b),n) There is another half , So we just need to enumerate to sqrt(b), Then find the number of factors , When enumerating to a factor , Another factor is found , If i * i It happens to be n, Then there is only one .

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll isPrime(ll n) {
    
    ll cnt = 0;
    for (ll i = 1; i <= sqrt(n); ++i) {
    
        if (n % i == 0) {
    
            if (i * i != n) {
    
                cnt += 2;
            } else {
    
                cnt++;
            }
        }
    }
    return cnt;
}
void solve() {
    
    ll b;
    scanf("%lld", &b);
    printf("%lld\n", isPrime(b));
}
int main() {
    
    solve();
    return 0;
}