UVa 11105 semi-prime H-numbers

方法：素数筛选

素数筛选法的推广。先求出H-primes， 可以证明得到，任意两个H-prime相乘，结果都是semi-prime H-number。求出范围内所有semi-primes，预处理前缀和即可。

(这里用到一种O(n) 素数筛选的方法）

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
  using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

set<ll> ans;
vector<ll> primes;
const int maxn = 25e4;
bitset<maxn+1> vis(0);
int cnt[4*maxn+1] = {0};
void init()
{

    repn(i, maxn)
    {
        if (!vis[i])
            primes.pb(i);
        for (auto p : primes)
        {
            ll nxt = i + p + 4 * p * i;
            if (nxt > maxn) break;
            vis[nxt] = true;
            if ((i-p)%(4*p+1) == 0) break;
        }
    }
    for (auto i : primes)
        for (auto p : primes)
        {
            ll nxt = i + p + 4 * p * i;
            if (nxt > maxn) break;
            cnt[nxt*4+1] = true;
        }
    repn(i, 4*maxn) cnt[i] += cnt[i-1];
}





int main()
{
    init();
    int n;
    while (cin >> n && n)
    {
        cout << n << " " << cnt[n] << newline;
    }
    
}