Semi-prime H-numbers POJ

Semi-prime H-numbers

POJ - 3292

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of4n+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 theH-numbers. For this problem we pretend that these are the only numbers. TheH-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. AnH-number h is H-prime if it is not the unit, and is the product of twoH-numbers in only one way: 1 × h. The rest of the numbers areH-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 twoH-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 threeH-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 857890

Sample Output

21 085 5789 62

题意：定义伪素数n，（n-1）%4==0。伪素数相乘是超级伪素数，但是三个伪素数相乘得到的不是超级伪素数。求从1到N一共有多少个超级伪素数。

分析：筛法的简单变形，将筛素数的思想用上就可以了。难点主要在于控制偶数个伪素数相乘，这里用一个VIS【MAXN】数组来控制，一共有很多总情况，但是我们不必考虑数量，只需要考虑两个数相乘是是偶数个伪素数还是奇数个伪素数在相乘就行了，所以用1代表偶数个，用-1代表奇数个，0代表伪素数。。。合为偶数就能成为超级伪素数。。

收获：筛法的技巧，用s这个数组的-1,1,0值代表各种状态，对思维的训练还是有的

代码：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int maxn=1000001;int s[maxn],t[maxn],vis[maxn];int m;void Init(){    memset(s,0,sizeof(s));    memset(vis,0,sizeof(vis));    memset(t,0,sizeof(t));    for(int i=5; i<maxn; i+=4)    {        for(int j=5; j<maxn; j+=4)        {            if(i*j>=maxn)            {                break;            }            if(!s[i]&&!s[j])            {                s[i*j]=1;            }            else            {                s[i*j]=-1;            }        }    }    int sum =0;    for(int i=5; i<maxn; i++)    {        if(s[i]==1)        {            sum++;        }        t[i]=sum;    }    return ;}int main (){    Init();    while(scanf("%d",&m)!=EOF&&m)    {        printf("%d %d\n",m,t[m]);    }    return 0;}