poj 3292 Semi-prime H-numbers(筛法~)

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8736		Accepted: 3825

Description

This problem is based on an exercise of David Hilbert, who pedagogically(教学法上) suggested that one study the theory of 4n+1 numbers.
Here, we do only a bit of that.

An H-number is a positive(积极的) 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(乘法).

As with regular integers(整数), we partition(分割) 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(投入) 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(包括的),
separated by one space in the format shown in the sample(样品).

Sample Input

21
85
789
0

Sample Output

21 0
85 5
789 62

题意：先定义一些数：1、H-number：可以写成4k+1的数，k为整数2、H-primes:只能分解成1*本身，不能分解成其他的H-number3、H-semi-primes：能够恰好分解成两个H-primes的乘积，且只能是两个数的乘积。输出1-N之间有多少个H-semi-primes。解法：类似于筛法求素数。先筛出H-primes，然后再枚举每两个H-primes的乘积，筛出H-semi-primes。最后统计1-Max之间的H-semi-primes的个数，保存在数组中。

#include<stdio.h>

int h[1000001];
int n;
void slove()
{
int i,j,s,count;
for( i = 5;i <=1000001; i+=4)
for( j = 5; j <= 1000001; j+=4)
{
s=i*j;
if(s>1000001) break;
if(h[i]==0&&h[j]==0)
{
h[s] = 1;
}
else h[s]=-1;
}
count = 0;
for(i=1;i<=1000001;i++)
{
if(h[i]==1)
{
h[i]=++count;
}
else h[i] = count;
}
}
int main()
{

int i,j;

//    for( i = 5;i<=100; i++)
//    if(h[i]==1)
//        printf("%d ",h[i]);
slove();
while(~scanf("%d",&n),n)
{

//
//        int sum=0;
//        for(i=5;i<=n;i++)//这样会超时  所以 ……
//            if(h[i]==1)
//              sum++;
printf("%d %d\n",n,h
);
}
return 0;
}