POJ 3292 Semi-prime H-numbers（打表）

题意大体是：规定如果一个数是由1,5,9,13,17,21、、、这么一个等差数列中的任意两项组成的乘积组成的数称之为H-numbers。然后给你一个数输出比这个数字小的H-numbers有多少个、、、（貌似不能是乘1）。

类似的一个打表，数据范围也不广，可以水过啊、、、

Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6888 Accepted: 2943

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 theonly 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

#include <stdio.h>#include <string.h>#include <cmath>#include <iostream>#include <stdlib.h>#include <map>using namespace std;int f[1001000];void _printf(){    int i, j;    memset(f , 0 , sizeof(f));    for(i = 5; i <= 1000; i+= 4)    {        if(f[i] == 0)        {            for(j = i; i*j < 1000002; j+=4)            {                f[i*j] = f[i]+f[j]+1;            }        }    }    for(i = 25; i < 1000002; i++)    {        if(f[i] == 1)            f[i] = f[i-1]+1;        else            f[i] = f[i-1];    }}int main(){    int n;    _printf();    while(cin >>n)    {        if(!n)            break;        cout <<n<<' '<<f[n]<<endl;    }    return 0;}