POJ 题目3292 Semi-prime H-numbers（打表，数学）

Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7630 Accepted: 3297

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

Source

Waterloo Local Contest, 2006.9.30

ac代码

#include<stdio.h>#include<string.h>int h[1000010];void fun(){int i,j,num,cot=0;for(i=5;i<1000010;i+=4){for(j=5;j<1000010;j+=4){num=i*j;if(num>=1000010)break;if(!h[i]&&!h[j])h[num]=1;elseh[num]=-1;}}for(i=1;i<1000010;i++){if(h[i]==1)cot++;h[i]=cot;}}int main(){int n;fun();while(scanf("%d",&n)!=EOF,n){printf("%d %d\n",n,h[n]);}}