This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it‘s the product of three H-primes.

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <algorithm>
 5 #include <cstring>
 6 using namespace std;
 7 
 8 const int MAXN = 1000001;
 9 bool is_prime[MAXN + 5];
10 int x;
11 vector<int> res;
12 void init()
13 {
14     vector<int> prime;
15     memset(is_prime, 1, sizeof(is_prime));
16     for (int i = 5; i <= MAXN; i += 4)
17     {
18         if (is_prime[i])
19         {
20             prime.push_back(i);
21             for (int j = 5; i * j <= MAXN; j += 4)
22             {
23                 is_prime[i * j] = false;
24             }
25         }
26     }
27     for (int i = 25; i <= MAXN; i += 4)
28     {
29         if (is_prime[i])
30             continue;
31         int tmp = i;
32         int cnt = 0;
33         for (int j = 0; prime[j] * prime[j] <= tmp; j++)
34         {
35             if (cnt > 2)
36                 break;
37             while (tmp % prime[j] == 0)
38             {
39                 cnt++;
40                 tmp /= prime[j];
41             }
42         }
43         if (tmp != 1)
44             cnt++;
45         if (cnt <= 2)
46             res.push_back(i);
47     }
48 }
49 
50 int main()
51 {
52     init();
53     while (scanf("%d", &x), x)
54     {
55         printf("%d %d\n", x, upper_bound(res.begin(), res.end(), x) - res.begin());
56     }
57     return 0;
58 }