PAT Advanced 1069 The Black Hole of Numbers (20分)
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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 (.
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
这道题主要考察了字符串的转换,我们使用转换字符串的方式,可以快速得到答案,
使用do while循环,避免了0或刚开始是6767不输出情况
#include <iostream> #include <algorithm> using namespace std; int main() { int N; scanf("%d", &N); string str, rev; do { str = to_string(N); while(str.length() != 4) str = "0" + str; sort(str.begin(), str.end()); rev = str; reverse(rev.begin(), rev.end()); N = stoi(rev) - stoi(str); printf("%s - %s = %04d ", rev.c_str(), str.c_str(), N); }while(N != 0 && N != 6174); return 0; }
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