POJ 3292 Semi-prime H-numbers （數論）

Semi-prime H-numbers Time Limit: 1000MS Memory Limit: 65536K

Description

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 areH-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

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.

Output

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.

Sample Input

21 
85
789
0

題意：先定義一些數：

1、H-number：可以寫成4k+1的數，k為整數

2、H-primes:只能分解成1*本身，不能分解成其他的H-number

3、H-semi-primes：能夠恰好分解成兩個H-primes的乘積，且只能是兩個數的乘積。輸出1-N之間有多少個H-semi-primes。

解法：類似於篩法求素數。先篩出H-primes，然後再枚舉每兩個H-primes的乘積，篩出H-semi-primes。最後統計1-Max之間的H-semi-primes的個數，保存在數組中。

#include
#include
#include
const int Max = 1000010;
int H_semi_prime[Max];

void solve()
{
    int i, j;
    memset(H_semi_prime, 0, sizeof(H_semi_prime));
    for(i = 5; i <= 1000001; i += 4)
        for(j = 5; j <= 1000001; j += 4)
        {
            int product = i * j;
            if(product > Max)
                break;
            if(H_semi_prime[i] == 0 && H_semi_prime[j] == 0)
                H_semi_prime[product] = 1;
            else
                H_semi_prime[product] = -1;
        }
    int cnt = 0;
    for(i = 0; i <= 1000001; i++)
    {
        if(H_semi_prime[i] == 1)
            cnt++;
        H_semi_prime[i] = cnt;
    }
}

int main()
{
    int n;
    solve();
    while(~scanf("%d",&n) && n)
    {
         printf("%d %d\n",n, H_semi_prime[n]);
    }
    return 0;
}