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 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 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 hinclusive, separated by one space in the format shown in the sample.
Sample Input
21 85 789 0
Sample Output
21 0 85 5 789 62
解题思路:
暴力打表,从5开始枚举,步长为4:这样得到的枚举结果是H数
将H数两两相乘,如果乘数都为H质数,那么结果就是半H质数
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAXN = 1000010;
int N;
int is_primes[MAXN];
int tanses[MAXN];
void getTables() {
for (int i = 5; i <= MAXN; i+=4) {
for (int j = 5; j <= MAXN; j+=4) {
int mulres = i * j;
if(mulres > MAXN) break; //超出范围就退出
if (is_primes[i]==0 && is_primes[j]==0) { //如果两个都是质数
is_primes[mulres] = 1;
}
else {
is_primes[mulres] = -1; //为什么设置为-1?因为它已经是两个数的乘积,不满足条件
}
}
}
int tans = 0;
for (int k = 1; k <= MAXN; ++k) {
if (is_primes[k]==1) {
tans++;
}
tanses[k] = tans;
}
}
int main() {
memset(is_primes, 0, sizeof(is_primes));
getTables(); //打表
while (scanf("%d", &N) != EOF) {
if(N == 0) break;
printf("%d %d\n", N, tanses[N]);
}
system("PAUSE");
return 0;
}