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 (0,104).
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
解题思路:
将一个数的从大到小排序减去从小到大排序,值赋给这个数,依次循环直到得到6174或者0
这题的注意点是如果输入的数或者得到的值不足4位数的话,要用0补足
这题用string就可以解决,利用stoi和to_string函数互相转换
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
bool cmpMAx(char a, char b) {
return a > b;
}
bool cmpMin(char a, char b) {
return a < b;
}
string num;
int main() {
cin >> num;
int res = -1;
string maxsnum, minsnum;
while (true) {
if (num.length() < 4) {
for (int j = num.length(); j < 4; ++j) {
num += '0';
}
}
sort(num.begin(), num.end(), cmpMAx);
maxsnum = num;
sort(num.begin(), num.end(), cmpMin);
minsnum = num;
res = stoi(maxsnum) - stoi(minsnum);
printf("%04d - %04d = %04d\n", stoi(maxsnum), stoi(minsnum), res);
if(res == 0 || res == 6174) break;
num = to_string(res);
}
system("PAUSE");
return 0;
}