1069 The Black Hole of Numbers（20 分）
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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 61747641 - 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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174
Sample Input 2:
2222
Sample Output 2:
2222 - 2222 = 0000
题目大意：给定一个四位数，展示出每位数从大到小排列，与从小到大排列的差值，直到出现6174黑洞数停止。
//这道题目是简单的，但是还是遇见了一些问题：
AC：
#include#include #include #include #include using namespace std;int main(){ int n; cin>>n; int a[4],big,small,res=-1; int temp=0; while(res!=6174) { fill(a,a+4,0); while(n!=0) { a[temp++]=n%10; n/=10; //cout<<"kk"; } temp=0; sort(a,a+4);//默认从小到大排列 small=a[0]*1000+a[1]*100+a[2]*10+a[3]; big=a[3]*1000+a[2]*100+a[1]*10+a[0]; if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3]) { printf("%04d - %04d = 0000\n",big,small); return 0; } res=big-small; printf("%04d - %04d = %04d\n",big,small,res); n=res; } return 0;}
1.第一次忽略了temp=0；应该在使用过后将其赋值为0的；
2. 应该将a数组初始化为0，要不然下次会有影响的，比如在有0位的时候：
3.最后就是提交发现第0个测试点过不去，原来是因为相同数的时候只输出了0，而不是0000！