UVA 10539 - Almost Prime Numbers 素数打表

Almost prime numbers are the non-prime numbers which are divisible by only a single prime number.
In this problem your job is to write a program which finds out the number of almost prime numbers
within a certain range.
Input
First line of the input file contains an integer N (N ≤ 600) which indicates how many sets of inputs
are there. Each of the next N lines make a single set of input. Each set contains two integer numbers
low and high (0 < low ≤ high < 1012).
Output
For each line of input except the first line you should produce one line of output. This line contains
a single integer, which indicates how many almost prime numbers are within the range (inclusive)
low . . . high.
Sample Input
3
1 10
1 20
1 5
Sample Output
3
4
1

题意：给你 一个范围 a,b，找出这个范围中 满足 x = p^k (p为素数，k > 1)  的数的个数

题解： ab,范围是10的12次方 我们找出1e6内的素数 ，打表出所有可能形成 的数去重排序 ，每次二分找下标就好了

ans(b) - ans(a-1)就是答案

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<vector>
using namespace std ;
typedef long long ll;

const int N=;
ll a, b, prime[N + ], now[N+];
int H[N + ], cnt, scc;
void prime_table() {
    cnt = ;
    H[] = ;
    for(int i = ; i <= N ; i++) {
        if(!H[i]) {
            for(int j = i + i ; j <= N ; j += i) H[j] = ;
            prime[++cnt] = i;
        }
    }
    scc = ;
    for(int i =  ; i <= cnt ; i++) {
        for(ll j = prime[i] * prime[i] ; j <= 1e12 ; j *= prime[i]) {
            now[scc++] = j;
        }
    }
    sort(now, now + scc);
    scc = unique(now,now + scc) - now ;
}
ll solve(ll x) {
    ll ans = upper_bound(now,now + scc,x) - now;
    return ans;
}
int main() {
    prime_table();
    int T;
    scanf("%d",&T);
    while(T--) {
        scanf("%lld%lld",&a,&b);
        ll ans = solve(b) - solve(a-);
        printf("%lld\n", ans);
    }
    return ;
}

代码