A1069. The Black Hole of Numbers

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:

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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 bool cmp1(int a, int b){
 6     return a < b;
 7 }
 8 bool cmp2(int a, int b){
 9     return a > b;
10 }
11 void numSort(int n, int &r1, int &r2){
12     int temp[20];
13     int i = 0;
14     r1 = 0; r2 = 0;
15     do{
16         temp[i++] = n % 10;
17         n = n / 10;
18     }while(n != 0 || i < 4);
19     sort(temp, temp + i, cmp1);
20     for(int j = 0, P = 1; j < i; j++){
21         r1 = r1 + P * temp[j];
22         P = P * 10;
23     }
24     sort(temp, temp + i, cmp2);
25     for(int j = 0, P = 1; j < i; j++){
26         r2 = r2 + P * temp[j];
27         P = P * 10;
28     }
29 }
30 int main(){
31     int N, r1, r2, ans;
32     scanf("%d", &N);
33     numSort(N, r1, r2);
34     do{
35         ans = r1 - r2;
36         printf("%04d - %04d = %04d\n", r1, r2, ans);
37         numSort(ans, r1, r2);
38     }while(ans != 6174 && ans != 0);
39     cin >> N;
40     return 0;
41 }
View Code

 

总结:

1、注意在int转换为num[ ]数组时,如果不够四位,应补全成四位,否则答案会出错。(15应转换为0015和1500,而不是15和50)。

posted @ 2018-02-04 14:07  ZHUQW  阅读(183)  评论(0编辑  收藏  举报