首页 > 代码库 > POJ 3292 Semi-prime H-numbers(数)
POJ 3292 Semi-prime H-numbers(数)
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
0
Sample Output
21 0
85 5
789 62
题意 所有可以表示为4*k+1(k>=0)的数都称为“H数” 而在所有“H数”中只能被1和自身整除的H数称为“H素数“ 能表示成两个”H素数“积的数又称为”Semi-prime H数“
输入n 求1到n之间有多少个”Semi-prime H数“;
方法 先打个H素数表 再用H素数表中的数依次相乘 得到的数都标记 再用一个数组保存每个数以内的标记数 输入n后直接读数组就行了
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int N=1000001; int vis[N],hp[N],ans[N],n; int main() { int num=0,m=sqrt(N+0.5); for(int i=5;i<=m;i+=4) { if(vis[i]==0) for(int j=i*i;j<=N;j+=i) vis[j]=1; } for(int i=5;i<N;i+=4) if(!vis[i]) hp[++num]=i; memset(vis,0,sizeof(vis)); for(int i=1;hp[i]*hp[i]<=N;++i) for(int j=i;hp[i]*hp[j]<=N;++j) ++vis[hp[i]*hp[j]]; num=0; for(int i=1;i<N;++i) { if(vis[i]>=1) ++num; ans[i]=num; } while(scanf("%d",&n),n) printf("%d %d\n",n,ans[n]); return 0; }