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Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9457 Accepted: 4202

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 are H-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

题目不难，只是借这个题写一下艾式筛法。

艾式筛法求素数代码（对于每一个素数，它的N倍一定不是素数）：

for(int i=2;i<maxn;i++){    if(!vis[i])    {        vis[i]=1;        prime[cnt++]=i;        for(int j=i*i;j<maxn;j*=i)            vis[j]=1;    }}

附上AC代码：

#include<iostream>using namespace std;const int maxn=1000000+5;int cnt[maxn];int a;void is_Hprime(){    for(int i=5;i<maxn;i+=4)    {        for(int j=5;j<maxn;j+=4)        {            if(i*j>maxn)break;            if(!cnt[i]&&!cnt[j])            {                cnt[i*j]=1;            }            else cnt[i*j]=-1;        }    }    int num=0;    for(int i=25;i<maxn;i++)    {        if(cnt[i]==1)            num++;        cnt[i]=num;    }}int main(){    is_Hprime();    while(cin>>a,a)        cout<<a<<' '<<cnt[a]<<endl;}