For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner
we will soon end up at the number 6174 -- the "black hole" of 4-digit numbers. This number is named Kaprekar Constant.

For example, start from 6767, we'll get:

7766 - 6677 = 1089

9810 - 0189 = 9621

9621 - 1269 = 8352

8532 - 2358 = 6174

7641 - 1467 = 6174

... ...

Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range (0, 10000).

Output Specification:

If all the 4 digits of N are the same, print in one line the equation "N - N = 0000". Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.

Sample Input 1:

6767

Sample Output 1:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000
#include <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"string.h"
#include"algorithm"
using namespace std;
bool compareChar(char c1, char c2){
if(c1<c2)
return false;
return true;
}
char * stringMinus(char *s1, char *s2){
for(int i=3;i>=0;i--){
if(s1[i]>=s2[i]){
s1[i] = '0'+(s1[i]-s2[i]);
}
else{
s1[i] = '0'+(s1[i]-s2[i]+10);
if(i>0)
s1[i-1] -= 1;
}
}
return s1;
} int main()
{
int n;
char s[5]="0000";
char incr[5];
scanf("%d",&n);
int i=0;
while(n){
int x = n%10;
s[3-i] = '0'+x;
n/=10;
i++;
}
while(strcmp(s,"0000")!=0){
sort(s,s+4);
strcpy(incr,s); sort(s,s+4,compareChar); printf("%s - %s = ",s,incr);
printf("%s\n",stringMinus(s,incr));
if(strcmp(s,"6174")==0){
break;
}
}
return 0;
}

1069. The Black Hole of Numbers (20)的更多相关文章

  1. 1069. The Black Hole of Numbers (20)【模拟】——PAT (Advanced Level) Practise

    题目信息 1069. The Black Hole of Numbers (20) 时间限制100 ms 内存限制65536 kB 代码长度限制16000 B For any 4-digit inte ...

  2. PAT 甲级 1069 The Black Hole of Numbers (20 分)(内含别人string处理的精简代码)

    1069 The Black Hole of Numbers (20 分)   For any 4-digit integer except the ones with all the digits ...

  3. 1069 The Black Hole of Numbers (20分)

    1069 The Black Hole of Numbers (20分) 1. 题目 2. 思路 把输入的数字作为字符串,调用排序算法,求最大最小 3. 注意点 输入的数字的范围是(0, 104), ...

  4. PAT 1069. The Black Hole of Numbers (20)

    For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in ...

  5. PAT Advanced 1069 The Black Hole of Numbers (20) [数学问题-简单数学]

    题目 For any 4-digit integer except the ones with all the digits being the same, if we sort the digits ...

  6. PAT (Advanced Level) 1069. The Black Hole of Numbers (20)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  7. PAT甲题题解-1069. The Black Hole of Numbers (20)-模拟

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789244.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  8. 【PAT甲级】1069 The Black Hole of Numbers (20 分)

    题意: 输入一个四位的正整数N,输出每位数字降序排序后的四位数字减去升序排序后的四位数字等于的四位数字,如果数字全部相同或者结果为6174(黑洞循环数字)则停止. trick: 这道题一反常态的输入的 ...

  9. PAT 1069 The Black Hole of Numbers

    1069 The Black Hole of Numbers (20 分)   For any 4-digit integer except the ones with all the digits ...

随机推荐

  1. view的setTag() 和 getTag()应用 (转)

    原文地址:http://www.cnblogs.com/qingblog/archive/2012/07/03/2575140.html View中的setTag(Onbect)表示给View添加一个 ...

  2. FastDFS简介

    一.FastDFS概述: FastDFS是一个开源的轻量级分布式文件系统,他对文件进行管理,功能包括:文件存储.文件同步.文件访问(文件上传.下载)等,解决了大容量存储和负载均衡的问题,高度追求高性能 ...

  3. java判断字符串是否为数字或中文或字母

     个人认为最好的方法 *各种字符的unicode编码的范围:     * 汉字:[0x4e00,0x9fa5](或十进制[19968,40869])     * 数字:[0x30,0x39](或十进制 ...

  4. 【转载】C++中public,protected,private访问

    第一:private, public, protected 访问标号的访问范围. 假如我们约定: 类内部-----指的是当前类类型的定义中,以及其成员函数的声明和定义中: 类外部-----指的是不在当 ...

  5. iOS10 UI教程视图的绘制与视图控制器和视图

    iOS10 UI教程视图的绘制与视图控制器和视图 iOS10 UI视图的绘制 iOS10 UI教程视图的绘制与视图控制器和视图,在iOS中,有很多的绘图应用.这些应用大多是在UIView上进行绘制的. ...

  6. spring优化使用

    1.bean由框架填充,避免手写优化代码. 2.view的展示通过配置或注解实现最优化使用架构. 待续...

  7. uva 12003 分块

    大白上的原题,我就练练手... #include <bits/stdc++.h> using namespace std; typedef long long ll; ; ; ll blo ...

  8. js 获取浏览器高度和宽度值(多浏览器)

    IE中: document.body.clientWidth ==> BODY对象宽度 document.body.clientHeight ==> BODY对象高度 document.d ...

  9. git 学习笔记4--.gitignore

    很多时候,我们都不希望非源码的文件加入到repository管理. 这时,.gitignore文件就上场了. ignore规则 所有空行或者以注释符号 # 开头的行都会被 Git 忽略. 可以使用标准 ...

  10. java读取utf8配置文件乱码

    email.properties文件如果以ISO-8859-1编码,那么以下的java代码读取中文不会乱码,因为eclipse下中文都被翻译成/u... //in Conf.javaPropertie ...