Poj3292 Semi-prime H-numbers, pojsemi-prime

Poj3292 Semi-prime H-numbers, pojsemi-prime

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:10194		Accepted:4533

Description

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory4n + 1Numbers. Here, we do only a bit of that.

AnH-Number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... areH-Numbers. For this problem we pretend that these areOnlyNumbers.H-Numbers are closed under multiplication.

As with regular integers, we partitionH-Numbers into units,H-Primes, andH-Composites. 1 is the only unit.H-NumberHIsH-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 fewH-Composites are: 5x5 = 25, 5x9 = 45, 5x13 = 65, 9x9 = 81, 5x17 = 85.

Your task is to count the numberH-Semi-primes.H-Semi-prime isH-Number which is the product of exactly twoH-Primes. The twoH-Primes may be equal or different. In the example above, all five numbers areH-Semi-primes. 125 = 5 × 5 × 5 is notH-Semi-prime, because it's the product of threeH-Primes.

Input

Each line of input containsH-Number ≤ 1,000,001. The last line of input contains 0 and this line shocould not be processed.

Output

For each inputtedH-NumberH, Print a line statingHAnd the numberH-Semi-primes between 1 andHRandom Sive, separated by one space in the format shown in the sample.

Sample Input

21 857890

Sample Output

21 085 5789 62

Source

Waterloo Local Contest, 2006.9.30 if a number $ I $ is the prime number required by the question, then $5 * I + 4 * I * x $ is not the prime number required by the question, use the sieve prime number method to screen the number of non-prime numbers, and then obtain and maintain the prefix of the statistical answer.

#include<cstdio>#include<cmath>#include<iostream>#include<cstring>using namespace std;const int MAXN=1e6+10;const int limit=1e6+10;inline int read(){    char ch=getchar();int f=1,x=0;    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int Prime[MAXN],vis[MAXN],sum[MAXN],ans[MAXN];int tot=0;int main(){    for(int i=5;i<=limit;i+=4)    {        if(vis[i])  continue;        Prime[++tot]=i;        for(int j=i*5;j<=limit;j+=i*4)            vis[j]=1;    }    for(int i=1;i<=tot;i++)        for(int j=1;j<=i&&Prime[i]*Prime[j]<=limit;j++)            ans[Prime[i]*Prime[j]]=1;    for(int i=1;i<=limit;i++)   sum[i]=sum[i-1]+ans[i];    int h;    while(scanf("%d",&h)&&h)        printf("%d %d\n",h,sum[h]);    return 0;  }