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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;
}
URAL 2070 Interesting Numbers (找规律)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5796882.html
