| 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-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>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[n]);}return 0;
}