Interesting Numbers URAL - 2070 (规律）

最新推荐文章于 2020-07-05 20:44:31 发布

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博客讲述Nikolay和Asya对有趣整数的不同定义，Nikolay认为质数有趣，Asya认为正除数数量为质数的整数有趣，他们称两人看法一致的整数为满足的整数。他们要研究区间[L; R]内的数，需计算该区间内满足条件的整数数量，思路是减去不符合的。

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Interesting Numbers

URAL - 2070

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;
}