Semi-prime H-numbers 筛选法

Semi-prime H-numbers

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 6868Accepted: 2927

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 857890

Sample Output

21 085 5789 62

这个题目的意思 H-numbers 是4n+1的数，H-primes是指那些因子只有自身和1的 H-numbers 相乘得到的数字，如5

=1*5算是，但是25=5*5=1*25，就不算了，他让我们求出的数Semi-prime H-numbers是指由H-primes相乘得到的数字，

举个例子来说，45=5*9，因为5和9都只有1*5,9=1*9组成，这时候可能有人问9=3*3，不是素数啊，这个数的意思，它

的因子必须是H-numbers ，所以3不能算呗，所以45就是了，因为这个范围不大，所以筛选法就可以求出来了

那么这个题目的思路是啥的，我们要保证这个数的两个因子都符合条件，也就是说5*9=45，而45是这个数，但是45这

个数就不能算是了，所以我们不能选择45再作为因子了，我们用两个for循环逐层遍历，但是有人可能会问我们这样

能保证因子全是吗？，当然可以当我们遍历所以因子，列入当我们遍历到25时，我们就不能算了，因为在一开始5*5

就已经标记它不行了！！！！

还有一点我本来想求出求出那些数是就行了，但是还是过不了，tle,我们用每个范围对应的情况都求出就是了，所以

直接输出数组的值就好了

#include<iostream>#include<cstdio>#include<cstring>#define M 1000001int h[M+10];int count;void table(){    memset(h,0,sizeof(h));    int i,j,s;    for(i=5; i<=M; i=i+4)    {        for(j=5; j<=M; j=j+4)        {            s=i*j;            if(s>M)            {                break;            }            if(h[i]==0&&h[j]==0)                h[s]=1;            else                h[s]=-1;        }    }    count=0;    for(i=1;i<=M;i++)    {        if(h[i]==1)        {          count++;        }        h[i]=count;    }}int main(){    int i,j,k,n;    table();    while(scanf("%d",&n)!=EOF)    {        if(n==0)        break;       /* count=0;        for(i=1; i<=n; i++)        {            if(h[i]==1)           {               count++;           }        }        printf("%d %d\n",n,count);*/        printf("%d %d\n",n,h[n]);    }    return 0;}