POJ 3292 Semi-prime H-numbers

Semi-prime H-numbers

Time Limit: 1000MSMemory Limit: 65536K

Total Submissions: 7372Accepted: 3158

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 
85
789

0

Sample Output


21 0
85 5

789 62

题意：不得不说题意很难懂！！就是找出1到输入的，只能拆成两个4n+1相乘的数的个数！！

不过这个题还是很有意义的，它让我深刻的认识到了打表的强大，这是继N皇后之后遇到的又一个打表神题！！

AC代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define M 1000001using namespace std;int aa[M+1];int bb[M+1];int main(){    int i,j;    int n;    int t=0,tj=0;    memset(aa,0,sizeof aa);    for(i=5;i<=M;i+=4)        for(j=5;j<=M;j+=4)        {            if(i*j>M)                break;            if(aa[i]==0&&aa[j]==0)                aa[i*j]=1;            else aa[i*j]=-1;        }    int ans=0;    for(i=1;i<=M;i++)    {        if(aa[i]==1)        {            ans++;        }        bb[i]=ans;    }    while(~scanf("%d",&n)&&n)    {        printf("%d %d\n",n,bb[n]);    }    return 0;}