poj3292--Semi-prime H-numbers(数论篇2)
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Description
This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of4n+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 theH-numbers. For this problem we pretend that these are the only numbers. TheH-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. AnH-number h is H-prime if it is not the unit, and is the product of twoH-numbers in only one way: 1 × h. The rest of the numbers areH-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 twoH-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 threeH-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 857890
Sample Output
21 085 5789 62
Source
题目大意:给定1,4,9,13,,,,的数列叫做H-numbers,如果一个数仅能被两个H-H-semi-primes相乘(不能是1)得到,那么他就是H-semi-primes,问1到n内的H-semi-primes有多少个。
筛法求出来,vis初始为0.,代表它仅能有1*自身得到,为1代表是一个H-H-semi-primes数,为-1代表其他。
当两个i,j, vis[i] == 0 && vis[j] == 0 那么vis[i*j] = 1 ;
#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int vis[1100000] , a[1100000] , cnt ;void init(){ int i , j ; memset(vis,0,sizeof(vis)) ; cnt = 0 ; for(i = 5 ; i <= 1010000 ; i += 4) { for(j = 5 ; j <= i ; j+= 4) { if( i*j > 1010000 ) break ; if( !vis[i] && !vis[j] ) vis[i*j] = 1 ; else vis[i*j] = -1 ; } } memset(a,0,sizeof(a)) ; for(i = 1 ; i < 1010000 ; i++) { a[i] = a[i-1] ; if( vis[i] == 1 ) a[i]++ ; } return ;}int main(){ int n ; init() ; while( scanf("%d", &n) && n ) { printf("%d %d\n", n, a[n]) ; } return 0;}
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