POJ 3292 Semi-prime H-numbers （变形埃氏筛法）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7777 | Accepted: 3364 |
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
在4n+1域中不断的筛去H-semi、H-composite 最后剩下的就是H-prime 不过同时我们也得了H-semi
然后预处理下 即可
AC代码如下：
//
// POJ 3292 Semi-prime H-numbers
//
// Created by TaoSama on 2015-03-31
// Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e6 + 10;
int n, H_num[N], H_semi[N];
void seive() {
//在4n+1域中 0表示H-prime 1表示H-semi 2表示H-composite
for(int i = 5; i < N; i += 4) {
for(int j = 5; j < N; j += 4) {
long long t = 1LL * i * j;
if(t >= N) break;
if(H_num[i] == 0 && H_num[j] == 0) H_num[t] = 1;
else H_num[t] = 2;
}
}
for(int i = 1; i < N; ++i) {
H_semi[i] += H_semi[i - 1];
if(H_num[i] == 1) H_semi[i] ++;
}
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
seive();
while(cin >> n && n) {
cout << n << ' ' << H_semi[n] << endl;
}
return 0;
}