PAT-A-1069 The Black Hole of Numbers 【验证】

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 (0,10​4​​).

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

---

说实话，有点坑，说好了的给四位数，结果测试案例竟然有不够四位的，这个注意一点，不足补零

还有就是 输出一定要是4位数，%04d格式输出

#include <iostream>
#include <string>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
int main(){
	string data;
	cin>>data;
	char c=data[0];
	if(data[1]==c&&data[2]==c&&data[3]==c){
	  printf("%s - %s = 0000\n",data.c_str(),data.c_str());    //四位相同情况
	  return 0;
	}
	vector<int> stu;
	int size=data.size();
	for(int i=0;i<size;i++){
		stu.push_back(data[i]-'0');                      //添加数据
	}
	for(int l=0;l<4-size;l++){
		stu.push_back(0);                                //不足四位补零
	}
	int big,small;
	while(true){
		big=0,small=0;
		sort(stu.begin(),stu.end());                     //排序
		small=stu[0]*1000+stu[1]*100+stu[2]*10+stu[3];
		big  =stu[3]*1000+stu[2]*100+stu[1]*10+stu[0];
		int temp=big-small;
		printf("%d - %04d = %04d\n",big,small,temp);
		if(temp==6174) break;
		stu.clear();
		while(temp/10){                                 //重置数据
			stu.push_back(temp%10);
			temp/=10;
		}
		stu.push_back(temp);
		while(stu.size()!=4){                          //不足补零
			stu.push_back(0);
		}
	}
	

	return 0;
}