URAL 2070 Interesting Numbers (找规律)

题意：在[L, R]之间求：x是个素数，因子个数是素数，同时满足两个条件，或者同时不满足两个条件的数的个数。

析：很明显所有的素数，因数都是2，是素数，所以我们只要算不是素数但因子是素数的数目就好，然后用总数减掉就好。打个表，找找规律，你会发现，

这些数除外的数都是素数的素数次方，然后就简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define frer freopen("in.txt", "r", stdin)

#define frew freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e6 + 5;

const int mod = 1e8;

const char *mark = "+-*";

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int a[maxn];

vector<int> prime;

int main(){

    for(int i = 2; i <= sqrt(1000001+0.5); ++i) if(!a[i]){

        for(int j = i * i; j <= 1000001; j += i) a[j] = 1;

    }

    for(int i = 2; i <= 1000001; ++i) if(!a[i])  prime.push_back(i);

    LL x, y;

    while(cin >> x >> y){

        int cnt = 0;

        for(int i = 0; i < prime.size(); ++i){

            LL t = (LL)prime[i] * (LL)prime[i];

            int k = 2;

            while(t <= y){

                ++k;

                if(!a[k] && t >= x && t <= y)  ++cnt;

                t *= prime[i];

            }

        }

        LL ans = y - x - cnt + 1;

        cout << ans << endl;

    }

    return 0;

}