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.
思路
有两种思路,一种是将输入的数按照char数组的格式来,但是这样就要定义char数组的运算规则和排序规则,这么做比较麻烦;第二种方法是将整数转成整形数组,排序完了之后又转回整形进行运算,相比第一种方法要简单。选后者!
代码
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
bool compare(int a, int b)
{
return a > b;
}
int* intToArr(int a, int* arr){
for(int i = 3; i >= 0; i--){
arr[i] = a % 10;
a /= 10;
}
return arr;
}
int arrToInt(int* arr){
int ans = 0;
for(int i = 3; i >= 0; i--){
ans += arr[i] * pow(10, 3 - i);
}
return ans;
}
int main(){
int input, third, * arr = new int[5];
scanf("%d", &input);
third = input;
while(true){
arr = intToArr(third, arr);
int first, second;
sort(arr, arr + 4, compare);
first = arrToInt(arr);
sort(arr, arr + 4);
second = arrToInt(arr);
third = first - second;
printf("%04d - %04d = %04d\n", first, second, third);
if(third == 0 || third == 6174)
break;
}
delete[] arr;
return 0;
}