Interesting Numbers
Nikolay and Asya investigate integers together in their spare time. Nikolay thinks an integer is interesting if it is a prime number. However, Asya thinks an integer is interesting if the amount of its positive divisors is a prime number (e.g., number 1 has one divisor and number 10 has four divisors).
Nikolay and Asya are happy when their tastes about some integer are common. On the other hand, they are really upset when their tastes differ. They call an integer satisfying if they both consider or do not consider this integer to be interesting. Nikolay and Asya are going to investigate numbers from segment [ L; R] this weekend. So they ask you to calculate the number of satisfying integers from this segment.
Input
In the only line there are two integers L and R (2 ≤ L ≤ R ≤ 10 12).
Output
In the only line output one integer — the number of satisfying integers from segment [ L; R].
Example
| input | output |
|---|---|
3 7 |
4 |
2 2 |
1 |
77 1010 |
924 |
思路：把不符合的减去就行了
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <string>
#include <functional>
#include <cstdio>
#include <cmath>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
using namespace std;
#define ll long long
#define F first
#define S second
#define p_b push_back
#define m_p make_pair
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int M = 1e7 + 7;
const int N = 1e6 + 100;
ll prime[N];
int isprime[N];
void getprime() {
memset(prime, 0, sizeof(prime));
int n = 1e6+20;
for (int i = 2; i <= n; ++i) {
if (!prime[i]) prime[++prime[0]] = i, isprime[i] = 1;
for (int j = 1; j <= prime[0] && prime[j] <= n/i; ++j) {
prime[prime[j]*i] = 1;
if (i % prime[j] == 0) break;
}
}
}
ll sum(ll x) {
ll not_is = 0;
for (int i = 1; ; ++i) {
ll y = prime[i]*prime[i];
int num = 3;
if (y > x) break;
while (y <= x) {
if (isprime[num])
++not_is;
y *= prime[i];
++num;
}
}
return x - not_is;
}
int main() {
getprime();
//printf ("%d\n", prime[0]);
ll l, r;
scanf ("%lld %lld", &l, &r);
printf ("%lld\n", sum(r) - sum(l-1));
return 0;
}