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 in nonincreasing 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 diference. 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

题目分析

已知一个最多有4位的整数N，对N降序排列-对N升序排列=下一次的整数N，循环处理直到N=6174为止，打印其处理过程（特殊情况：如果N的4位数字相同，打印并退出处理）

解题思路

1. 用字符串接收整数s，并用0左边补齐整数到4位
2. a=s,b=s,a降序排列后转为数字an，b升序排列转为数字bn
3. bn-an即为下次处理的整数

易错点

1. 若输入的数字为6174，需要打印7641 - 1467 = 6174（否则测试点5错误）（建议使用do...while可以省去针对开始输入为6174的单独处理）

知识点

1. 字符串中字符排序
   
   sort(s.begin(),s.end(),cmp);
2. 利用字符串操作对齐整数
   
   2.1 右对齐。如：4位对齐，输入1，要求得到"0001"；输入11，要求得到"0011"
   
   s.insert(0,4-s.length,'0');
   
   2.2 左对齐。4位对齐，输入1，要求得到"1000"，输入11，要求得到"1100"
   
   s.insert(s.length,4-s.length,'0');

Code

Code 01

#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(char a, char b) {
	return a>b;
}
int main(int argc, char * argv[]) {
	string s,a,b;
	cin>>s;
	s.insert(0,4-s.length(),'0');
	do {
		a=s,b=s;
		sort(a.begin(),a.end(),cmp);
		sort(b.begin(),b.end());
		int res = stoi(a)-stoi(b);
		s=to_string(res);
		s.insert(0,4-s.length(),'0');
		cout<<a<<" - "<<b<<" = "<<s<<endl;
	} while(s!="6174"&&s!="0000");

	return 0;
}