| Time Limit: 1000MS | Memory Limit: 65536K | |
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 85 789 0
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-number
3、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>
#include<string.h>
#include<math.h>
const int Max = 1000010;
int H_semi_prime[Max];
void solve()
{
int i, j;
memset(H_semi_prime, 0, sizeof(H_semi_prime));
for(i = 5; i <= 1000001; i += 4)
for(j = 5; j <= 1000001; j += 4)
{
int product = i * j;
if(product > Max)
break;
if(H_semi_prime[i] == 0 && H_semi_prime[j] == 0)
H_semi_prime[product] = 1;
else
H_semi_prime[product] = -1;
}
int cnt = 0;
for(i = 0; i <= 1000001; i++)
{
if(H_semi_prime[i] == 1)
cnt++;
H_semi_prime[i] = cnt;
}
}
int main()
{
int n;
solve();
while(~scanf("%d",&n) && n)
{
printf("%d %d\n",n, H_semi_prime[n]);
}
return 0;
}