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http://poj.org/problem?id=3292
题意:H-number为mod 4=1的数
H-prim为除了1和本身外不能被任何H-number整除的数
H-semi-prime为两个H-prim的乘积
求1-h有多少个H-semi-prime;
思路:打表
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <stack>
#include <vector>
#include <map>
using namespace std;
#define N 1000020
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))
typedef long long LL;
int Hprim[N];
void Init ()
{
for (int i=5; i<N; i+=4)
{
for (int j=5; j<N; j+=4)
{
if (i*j>=N) break;
if (!Hprim[i] && !Hprim[j] && i*j%4==1)
Hprim[i*j] = 1;///<span style="font-family: KaiTi_GB2312;font-size:18px; line-height: 26px;">H-semi-prime</span>
else Hprim[i*j] = -1;
}
}
int cnt = 0;
for (int i=5; i<N; i++)
{
if (Hprim[i]==1) cnt++;
Hprim[i] = cnt;
}
}
int main ()
{
int n;
Init();
while (scanf ("%d", &n), n)
{
printf ("%d %d\n", n, Hprim[n]);
}
return 0;
}