Nastya Studies Informaticstime limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.
We define a pair of integers (a, b) good, if GCD(a, b) = x and LCM(a, b) = y, where GCD(a, b) denotes the greatest common divisor of a and b, and LCM(a, b) denotes the least common multiple of a and b.
You are given two integers x and y. You are to find the number of good pairs of integers (a, b) such that l ≤ a, b ≤ r. Note that pairs (a, b) and (b, a) are considered different if a ≠ b.
Input
The only line contains four integers l, r, x, y (1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ y ≤ 109).
Output
In the only line print the only integer — the answer for the problem.
Examplesinput
Copy
1 2 1 2
output
Copy
2
input
Copy
1 12 1 12
output
Copy
4
input
Copy
50 100 3 30
output
Copy
0
Note
In the first example there are two suitable good pairs of integers (a, b): (1, 2) and (2, 1).
In the second example there are four suitable good pairs of integers (a, b): (1, 12), (12, 1), (3, 4) and (4, 3).
In the third example there are good pairs of integers, for example, (3, 30), but none of them fits the condition l ≤ a, b ≤ r.
给你四个数l，r，a，b，问在l到r的范围内有多少对数（两个数不能相同，顺序可以不同）满足gcd(x,y)=a,lcm(x,y)=b
枚举b的因子个数，再看这些因子每每两个的最大公约数是否等于a，等于a满足条件情况加一
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e3 + ;
typedef long long ll;
ll gcd( ll p, ll q ) {
if( p == ) {
return q;
}
if( q == ) {
return p;
}
return gcd( q, p%q );
}
int main(){
std::ios::sync_with_stdio(false);
ll l, r, a, b;
//cout << gcd( 16, 4 ) << endl;
while( cin >> l >> r >> a >> b ) {
ll num = ;
vector<ll> e;
for( ll i = ; i * i <= b; i ++ ) {
if( b % i == ) {
e.push_back(i);
if( i != b/i ) {
e.push_back(b/i);
}
//debug(i),debug(b/i);
}
}
for( ll i = ; i < e.size(); i ++ ) {
for( ll j = ; j < e.size(); j ++ ) {
if( gcd( e[i], e[j] ) == a && e[i] * e[j] == a * b
&& e[i] >= l && e[i] <= r && e[j] >= l && e[j] <= r ) {
num ++;
}
}
}
cout << num << endl;
}
return ;
}
