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. Unity3D打Box游戏

    先学习一些基本的脚本实现: 1.动态创建物体.默认位置是(0,0)位置 GameObject goNew = GameObject.CreatePrimitive(PrimitiveType.Cube ...

  2. Android学习网站

    1 <老罗Android应用开发视频教程> http://www.mobiletrain.org/about/news/android_video2.html

  3. Android ImageCache图片缓存,使用简单,支持预取,支持多种缓存算法,支持不同网络类型,扩展性强

    本文主要介绍一个支持图片自动预取.支持多种缓存算法的图片缓存的使用及功能.图片较大需要SD卡保存情况推荐使用ImageSDCardCache. 与Android LruCache相比主要特性:(1). ...

  4. HDU 4812 D Tree 树分治+逆元处理

    D Tree Problem Description   There is a skyscraping tree standing on the playground of Nanjing Unive ...

  5. 【java基础】]IO流

    IO流 概念: 流的概念源于unix中管道(pipe)的概念,在unix中,管道是一条不间断的字节流,用来实现程序或进程间的通信,或读写外围设备,外部文件等 一个流,一定能够会有源和去向(目的地),他 ...

  6. Glow Android 优化实践

    了解 Glow 的朋友应该知道,我们主营四款 App,分别是Eve.Glow.Nuture和Baby.作为创业公司,我们的四款 App 都处于高速开发中,平均每个 Android App 由两人负责开 ...

  7. 期望+DP ZOJ 3929 Deque and Balls

    题目链接 题意:给你n个数,按照顺序依次放入一个双端队列(可放在头部,也可以放在尾部),求xi > xi+1的期望 * 2^n mod (1e9 +7) 分析:期望*2^n=出现这种排法的概率* ...

  8. 接口测试第二课(Fiddler实现APP抓包)

    Fiddler简介: Fiddler是强大且好用的Web调试工具之一,它能记录客户端和服务器的http和https请求,允许你监视,设置断点,甚至修改输入输出数 据. Fiddler 的运行机制其实就 ...

  9. 跳转页面&回到上一页

    1.php <?php header('Location:2.php'); //echo不能在其前面 echo '1'; 2.js <?php echo '2'; echo '<me ...

  10. ACM 无线网络覆盖

    无线网络覆盖 时间限制:3000 ms  |  内存限制:65535 KB 难度:3   描述 我们的乐乐同学对于网络可算得上是情有独钟,他有一个计划,那就是用无线网覆盖郑州大学. 现在学校给了他一个 ...