Humble Numbers HDU - 1058
Write a program to find and print the nth element in this sequence
InputThe 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.
OutputFor 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. 丑数
dp[i]=min(min(dp[a]*2,dp[d]*7),min(dp[b]*3,dp[c]*5));
if(dp[i]==dp[a]*2) a++;
if(dp[i]==dp[b]*3) b++;
if(dp[i]==dp[c]*5) c++;
if(dp[i]==dp[d]*7) d++;
输出的时候 注意13 12 11
#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[];
int main()
{
dp[]=;
int a=,b=,c=,d=;
for(int i=;i<=;i++)
{
dp[i]=min(min(dp[a]*,dp[d]*),min(dp[b]*,dp[c]*));
if(dp[i]==dp[a]*) a++;
if(dp[i]==dp[b]*) b++;
if(dp[i]==dp[c]*) c++;
if(dp[i]==dp[d]*) d++; }
int n;
while(~scanf("%d",&n)&&n)
{
if(n%==&&n%!=)
printf("The %dst humble number is %d.\n",n,dp[n]);
else if(n%==&&n%!=)
printf("The %dnd humble number is %d.\n",n,dp[n]);
else if(n%==&&n%!=)
printf("The %drd humble number is %d.\n",n,dp[n]);
else
printf("The %dth humble number is %d.\n",n,dp[n]); } return ;
}
Humble Numbers HDU - 1058的更多相关文章
- Humble Numbers HDU - 1058 2分dp
JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two ...
- hdu 1058 dp.Humble Numbers
Humble Numbers Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Subm ...
- HDOJ(HDU).1058 Humble Numbers (DP)
HDOJ(HDU).1058 Humble Numbers (DP) 点我挑战题目 题意分析 水 代码总览 /* Title:HDOJ.1058 Author:pengwill Date:2017-2 ...
- HDU 1058 Humble Numbers (DP)
Humble Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- hdu 1058:Humble Numbers(动态规划 DP)
Humble Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU 1058 Humble Numbers (动规+寻找丑数问题)
Humble Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- HDU - The number of divisors(约数) about Humble Numbers
Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence ...
- HDU 1058 Humble Number
Humble Number Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humbl ...
- HDOJ 1058 Humble Numbers(打表过)
Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The ...
随机推荐
- data:image/png;base64 上传图像将图片转换成base64格式
大家可能注意到了,网页上有些图片的src或css背景图片的url后面跟了一大串字符,比如: data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAkAAAAJ ...
- eclipse中.project文件和.classpath文件详解
一.概述.project是项目文件,项目的结构都在其中定义,比如lib的位置,src的位置,classes的位置..classpath的位置定义了你这个项目在编译时所使用的$CLASSPATH. 二. ...
- Oracle(限定查询1)
3.1.认识限定查询 例如:如果一张表中有100w条数据,一旦执行了“SELECT * FROM 表”语句之后,则将在屏幕上显示表中的全部数据行的记录,这样即不方便浏览,也有可能造成死机的问题出现,所 ...
- [原][osg][osgearth]倾斜摄影2.文件格式分析:OSGB
倾斜摄影三维模型格式包含:*.osgb,*.dae等 文件格式包含:*.xml, *.desc, *.lfp等 例如:LocaSpace Viewer软件把osgb分块模型文件建立索引生成一个lfp文 ...
- 学习笔记47—PhotoShop技巧
1.photoshop里怎么给画布画对角线? photoshop里给画布画对角线有二种方法: 1) 选直线工具 从一角拉向另一对角 就OK了 非常简单: 2) 选钢笔工具 鼠标先点击某一角 然后再点击 ...
- QT绘制饼图
QT版本:QT5.6.1 QT绘制饼图,出问题的代码如下 void DrawPieDialog::paintEvent(QPaintEvent *event) { float startAngle=0 ...
- MySQL学习(六)
1 注意 select cout(*) from 表名: 查询的就是绝对的行数,哪怕某一列所有字段全部为NULL,也计算在内.而select cout(列名) form 表名:查询的是该列不为null ...
- Lua报错:invalid key to 'next'
1.问题产生的原因是,在一个循环里对table中的元素先进行置空操作,再进行增加新元素的操作,就会报这个错误. 2.比如下面的例子:(当中间的函数足够复杂并进行封装了的情况下,不会留意到存在这个问题) ...
- 关于MySQL大量数据分页查询优化
select * form user id in(select id from user limit 1000000,10);
- makefile 里的 := , = , +=
:= 是在这行代码的时候,直接展开右边的变量. = 是在最终左边变量被使用的时候,才把右边的变量展开. https://stackoverflow.com/questions/10227598/wha ...