Poj 3292（筛法变形）

Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6750 Accepted: 2868

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 857890

Sample Output

21 085 5789 62

Source

Waterloo Local Contest, 2006.9.30

就是在它定义的数上做筛法就可以了，需要真正理解这个算法。

#include<cstdio>#include<iostream>#include<algorithm>#include<vector>#include<cstring>using namespace std;const int maxn = 1000000 + 5;const int INF = 2000000000;typedef  pair<int, int> P;typedef long long LL;vector<LL> h;int p[maxn];int sum[maxn];void pre(){    h.clear();    for(int i = 1;i < maxn;i+=4){        h.push_back(i);    }    memset(p,0,sizeof(p));    for(int i = 1;i < h.size();i++){        if(p[h[i]] == 0){            for(int j = i;h[i]*h[j] < maxn;j++){                if(p[h[j]] == 0)   p[h[i]*h[j]] = 1;                else p[h[i]*h[j]] = 2;            }        }    }    sum[1] = 0;    for(int i = 2;i < maxn;i++){        sum [i] = sum[i-1];        if(p[i] == 1) sum[i]++;    }}int main(){    pre();    int n;    while(scanf("%d",&n)){        if(n == 0) break;        printf("%d %d\n",n,sum[n]);    }    return 0;}