Humble Numbers

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 
 

Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. 

Write a program to find and print the nth element in this sequence 

 

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n. 
 

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output. 
 

Sample Input

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0
 

Sample Output

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.
题意:输出能被2或3或5或7整除的第n个数,注意题目的11,12,13后面是th。。。
 #include<bits/stdc++.h>
using namespace std;
int main() {
int n,m2=,m3=,m5=,m7=,i,temp1,temp2;
int a[];
a[]=;
for(i=; i<=; i++) {
a[i]=min(min(*a[m2],*a[m3]),min(*a[m5],*a[m7]));
if(a[i]==*a[m2]) m2++;
if(a[i]==*a[m3]) m3++;
if(a[i]==*a[m5]) m5++;
if(a[i]==*a[m7]) m7++;
}
while(cin>>n,n) {
if(n%==&&n%!=)
printf("The %dst humble number is %d.\n",n,a[n]);
else if(n%==&&n%!=)
printf("The %dnd humble number is %d.\n",n,a[n]);
else if(n%==&&n%!=)
printf("The %drd humble number is %d.\n",n,a[n]);
else
printf("The %dth humble number is %d.\n",n,a[n]);
}
return ;
}

HDU1058Humble Numbers的更多相关文章

  1. hdu1058Humble Numbers(动态规划)

    Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  2. Java 位运算2-LeetCode 201 Bitwise AND of Numbers Range

    在Java位运算总结-leetcode题目博文中总结了Java提供的按位运算操作符,今天又碰到LeetCode中一道按位操作的题目 Given a range [m, n] where 0 <= ...

  3. POJ 2739. Sum of Consecutive Prime Numbers

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050 ...

  4. [LeetCode] Add Two Numbers II 两个数字相加之二

    You are given two linked lists representing two non-negative numbers. The most significant digit com ...

  5. [LeetCode] Maximum XOR of Two Numbers in an Array 数组中异或值最大的两个数字

    Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum re ...

  6. [LeetCode] Count Numbers with Unique Digits 计算各位不相同的数字个数

    Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n. Examp ...

  7. [LeetCode] Bitwise AND of Numbers Range 数字范围位相与

    Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers ...

  8. [LeetCode] Valid Phone Numbers 验证电话号码

    Given a text file file.txt that contains list of phone numbers (one per line), write a one liner bas ...

  9. [LeetCode] Consecutive Numbers 连续的数字

    Write a SQL query to find all numbers that appear at least three times consecutively. +----+-----+ | ...

随机推荐

  1. PHP截取字符串 兼容utf-8 gb2312

    <?php function subString($string,$length,$append = false) { if(strlen($string) <= $length ) { ...

  2. WebBrowser里网页根据文字判断来点击链接 无Name及ID时

    uses ActiveX, ComObj, MSHTML; 根据连接文字点击连接- 一般情况下的连接 Procedure HTMLClinkByText(text:string;Wbr:TWebBro ...

  3. SQL Server 基础之《学生表-教师表-课程表-选课表》

    一.数据库表结构及数据 建表 CREATE TABLE Student ( S# INT, Sname ), Sage INT, Ssex ) ) CREATE TABLE Course ( C# I ...

  4. rspec的一些常见用法

    这里讲了如何安装rspec,安装使用rspec. 下面介绍一下rspec中常见的使用方法. 下面是一个最简单的测试用例,判断true是不是等于true,should_be是旧的用法,新用法推荐使用ex ...

  5. C# 自定义集合

    自定义类型 public class Product { public int Id { get; set; } // 自增ID public string Name { get; set; } // ...

  6. oracle 11g 修改默认监听端口1521

    OS: Oracle Linux Server release 5.7 DB: Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - ...

  7. DWR在Spring中应用

    这里以传递一个对象为例,来说明dwr在Spring中是怎么配置的. JSP页面: <script src='dwr/interface/instructionOuterService.js'&g ...

  8. golang的函数

    在golang中, 函数是第一类值(first-class object), 即函数可以赋值与被赋值. 换言之, 函数也可以作为ReceiverType, 定义自己的method. 实例: http. ...

  9. windows phone 8.1开发笔记之Toast

    Toast(吐司)是wp屏幕上端弹出来的临时通知,他会存在7秒钟的时间,可以快速定位到用户需要的位置(当然是由开发者设置的) 1.创建一个Toast now,需要将你的app设置支持Toast 否则即 ...

  10. CPU 材料学才是最顶级的学科

    cpu的物理组成3部分:逻辑部件.寄存器.控制部件 CPU具有以下4个方面的基本功能:数据通信,资源共享,分布式处理,提供系统可靠性 cpu处理4过程:提取.解码.执行.写回 http://baike ...