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POJ 3292 Semi-prime H-numbers

2018-06-17 21:33:14来源：未知阅读 ()

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

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

Sample Output

21 0
85 5
789 62

Source

Waterloo Local Contest, 2006.9.30

如果一个数$i$是题目要求的素数

那么$5*i+4*i*x$一定不是题目要求的素数

根据这个性质，利用筛素数的方法把不是素数的数筛去

然后再拿筛出来的数统计答案

前缀和维护一下

#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;  
}