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
#include
#include
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;
} |