Semi-prime H-numbers
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7059   Accepted: 3030

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

Source

 
 
仿照素数的埃氏筛选法即可
 
 #include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream> using namespace std; #define maxn 1000005 bool H[maxn];
int ans[maxn],ele[maxn];
int len = ; void init() { for(int i = ; i <= maxn - ; i++) {
H[i] = (i % == );
} for(int i = ; i * i <= maxn - ; i += ) {
if(!H[i]) continue;
for(int j = i; j * i <= maxn - ; j++) {
H[j * i] = ;
}
} for(int i = ; i <= maxn - ; i += ) {
if(H[i]) {
ele[len++] = i;
}
} for(int i = ; i < len && ele[i] * ele[i] <= maxn - ; i++) {
for(int j = i; j < len && ele[j] * ele[i] <= maxn - ; j++) {
if(ele[i] * ele[j] % == )
ans[ ele[i] * ele[j] ] = ;
}
} for(int i = ; i <= maxn - ; i++) {
ans[i] += ans[i - ];
}
} int main() {
// freopen("sw.in","r",stdin); init(); int x;
while(~scanf("%d",&x) && x) {
printf("%d %d\n",x,ans[x]);
} return ; }

POJ 3292的更多相关文章

  1. 【POJ 3292】 Semi-prime H-numbers

    [POJ 3292] Semi-prime H-numbers 打个表 题意是1 5 9 13...这样的4的n次方+1定义为H-numbers H-numbers中仅仅由1*自己这一种方式组成 即没 ...

  2. POJ 3292 Semi-prime H-numbers

    类似素数筛... Semi-prime H-numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6873 Accept ...

  3. Mathematics:Semi-prime H-numbers(POJ 3292)

      Semi-prime H-numbers 题目大意,令4n+1的数叫H数,H数素数x的定义是只能被x=1*h(h是H数),其他都叫合数,特别的,当一个数只能被两个H素数乘积得到时,叫H-semi数 ...

  4. POJ 3292 Semi-prime H-numbers (素数筛法变形)

    题意:题目比较容易混淆,要搞清楚一点,这里面所有的定义都是在4×k+1(k>=0)这个封闭的集合而言的,不要跟我们常用的自然数集混淆. 题目要求我们计算 H-semi-primes, H-sem ...

  5. Semi-prime H-numbers POJ - 3292 打表(算复杂度)

    题意:参考https://blog.csdn.net/lyy289065406/article/details/6648537 一个H-number是所有的模四余一的数. 如果一个H-number是H ...

  6. poj 3292 H-素数问题 扩展艾氏筛选法

    题意:形似4n+1的被称作H-素数,两个H-素数相乘得到H-合成数.求h范围内的H-合成数个数 思路: h-素数                                            ...

  7. 筛选法 || POJ 3292 Semi-prime H-numbers

    5,9,13,……叫H-prime 一个数能且仅能由两个H-prime相乘得到,则为H-semi-prime 问1-n中的H-semi-prime有多少个 *解法:vis初始化为0代表H-prime, ...

  8. POJ 3292:Semi-prime H-numbers 筛选数

    Semi-prime H-numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8216   Accepted: 3 ...

  9. Day7 - I - Semi-prime H-numbers POJ - 3292

    This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study th ...

随机推荐

  1. mysql颠覆实战笔记(五)--商品系统设计(二):定时更新商品总点击量

    继续回到沈老师的MYSQL颠覆实战,首先回顾下上一节课的内容,请大家会看下上节课写的存储过程. 打开prod_clicklog表, 我们只要把日期(不含时分秒)的部分存在数据库中, 如果同一日期有相同 ...

  2. 添加远程链接MySQL的权限

    mysql> grant 权限1,权限2,…权限n on 数据库名称.表名称 to 用户名@用户地址 identified by ‘连接口令’; 权限1,权限2,…权限n代表select,ins ...

  3. PowerDesigner中遍历物理模型中的所有表,检查表代码、字段代码

    '***************************************************************************** '文件:CheckCode4SqlServ ...

  4. SequoiaDB数据库集群部署

    一般在多机环境下部署数据库的集群模式是比较繁琐的,下面我来分享一个如何通过shell脚本的方式简单.方便地部署我们的集群. 首先,我们要给机器配置信任关系,这样我们就无需手动的输入密码来执行ssh和s ...

  5. Spark 大数据平台 Introduction part 2 coding

    Basic Functions sc.parallelize(List(1,2,3,4,5,6)).map(_ * 2).filter(_ > 5).collect() *** res: Arr ...

  6. [转]bat中的特殊字符,以及需要在bat中当做字符如何处理

    bat中的特殊字符,以及需要在bat中当做字符如何处理 批处理.Bat 中特殊符号的实际作用,Windows 批处理中特殊符号的作用: @ \\隐藏命令的回显. ~ \\在for中表示使用增强的变量扩 ...

  7. URL地址下载图片到本地

    package test.dao; import eh.base.dao.DoctorDAO; import eh.entity.base.Doctor; import junit.framework ...

  8. andriod ADB命令的使用

    android ADB命令的使用 ADB是一个 客户端-服务器端 程序, 其中客户端是你用来操作的电脑, 服务器端是android设备. 先说安装方法, 电脑上需要安装客户端. 客户端包含在sdk里. ...

  9. 从零开始学ios开发(十八):Storyboards(下)

    这篇我们完成Storyboards的最后一个例子,之前的例子中没有view之间的切换,这篇加上这个功能,使Storyboards的功能完整呈现.在Storyboards中负责view切换的东西叫做“s ...

  10. (C学习基础)一,CMD的使用

    以后考研要考数据结构,所以系统的整理一下C学习笔记.这里开发用VS2013  CMD资料在这里下载 首先就是DOS系统的CMD的使用,是不是觉得会命令行语句的人特别牛呢,有时候使用命令行总是事半功倍, ...