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 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 h inclusive, separated by one space in the format shown in the sample.
Sample Input
21 85 789 0Sample Output
21 0 85 5 789 62定义H数为模 4 余 1 的数。 H素数为在H数的范围内不含有 1 和它本身之外的其他因数的数。 H 半素数为最多只能由两个H素数相乘得到的 H 数，这两个 H 素数可能相同， 可能不同， 一个 H 半素数可以表示成多种 两H素数相乘。对于输入的 x， 求 1 到 x 之间 H 半素数的个数。
首先模拟筛法求素数打一个 H 素数表， 再打一个 H 半素数表。 最后对于每一个 x 询问二分求个数。
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int maxn = 1000010; int Hprime[maxn];//H 素数 int tot;//H 素数的个数 int hsp[maxn];//H 半素数 int tot1;//H 半素数的个数 bool vis[maxn];//H 半素数已经在 hsp 中存在的标记 //打表 void GetTab() { //求 H 素数 for(int i= 5; i< maxn; i+= 4) //只在 H 数中遍历 { if(Hprime[i] == 0 ) Hprime[tot++] = i; for(int j= 0; j< tot && i*Hprime[j]< maxn; j++) { Hprime[i*Hprime[j]] = 1; if(i % Hprime[j] == 0) break; } } //求 H 半素数 for(int i= 0; i< tot; i++) for(int j= 0; j<= i; j++) { int p = Hprime[i]*Hprime[j]; if(p >= maxn) break; if(vis[p]) continue;//去重 hsp[tot1++] = p; vis[p] = true; } sort(hsp, hsp+tot1); } int main () { GetTab(); while(1) { int x; scanf("%d", &x); if(x == 0) break; //二分求个数 int cnt = upper_bound(hsp, hsp+tot1, x) - hsp; printf("%d %d\n", x, cnt); } return 0; }
