poj 3292 Semi-prime H-numbers(筛法~)
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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 857890
Sample Output
21 085 5789 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;}
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