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:
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,+1.0E+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:
Sample Input 2:
2222
Sample Output 2:
要点:1.字符串排序
2.字符串补零
思路:先考虑特殊情况 当四位数字相同时 输出后直接退出-->升序==降序 即四位数字相同;
利用while循环输出,当满足设定条件退出循环--> while(1){......break;}
printf能够格式化输出-->printf("%0md",int)//输出m位整形数据,不足时用0补全
易错点:由于输入的N范围为(0,10000),所以对输入的字符串要预处理,用0补齐四位
示例:
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
void test()
{
string str;
cin >> str;
while (1)
{
while (str.length() < 4)
str.insert(0, "0");
string str1 = str;
sort(str1.begin(), str1.end());
string str2 = str1;
reverse(str2.begin(), str2.end());
if (str2 == str1)
{
cout << str << " - " << str << " = 0000\n";
return;
}
int num1 = stoi(str2);
int num2 = stoi(str1);
printf("%04d - %04d = %04d\n", num1, num2, num1 - num2);
str = to_string(num1 - num2);
if (num1 - num2 == 6174)
break;
}
}
int main()
{
test();
system("pause");
return 0;
}