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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, 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 difference. All the numbers must be printed as 4-digit numbers.
Sample Input 1:6767Sample Output 1:
7766 - 6677 = 1089 9810 - 0189 = 9621 9621 - 1269 = 8352 8532 - 2358 = 6174Sample Input 2:
2222Sample Output 2:
2222 - 2222 = 0000
#include<stdio.h>#include <iostream>#include <algorithm>#include <vector>#pragma warning(disable:4996)using namespace std;vector<int> num;bool cmp(int a, int b) {return a < b;}int main(void) {int n,result=-1;int temp;int high, low;cin >> n;temp = n;while (1) {num.clear();while (temp>0) {num.push_back(temp % 10);temp = temp / 10;}if (num.size() < 4)for (int i = num.size(); i < 4; i++)num.push_back(0);sort(num.begin(), num.end(), cmp);low = num[0] * 1000 + num[1] * 100 + num[2] * 10 + num[3];high = num[3] * 1000 + num[2] * 100 + num[1] * 10 + num[0];result = high - low;printf("%04d - %04d = %04d\n", high, low, result);if (result == 6174 || result == 0)break;temp = result;}return 0;}
1069. The Black Hole of Numbers (20)
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原文地址：http://www.cnblogs.com/zzandliz/p/5023216.html
