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
坑点1、数字都要四位的2、如果是判断下一个式子的差不等于上个结果,然后跳出的话,那么需要判断,是不是第一次输出。比如,输入6174,结果就等于6174,那么就会没输出,直接跳出。
#include <iostream> #include <algorithm> #include<string> #include <sstream> #include <iomanip> using namespace std; int a1[]; int a2[]; int bb[]; bool cmp1(int a,int b) { return a>b; } bool cmp2(int a,int b) { return a<b; } int main() { string n; int i; while(cin>>n) { int tt,c1,c2; stringstream ss1; ss1<<n; ss1>>tt; i=; bool fir=true; while(true) { string ss; stringstream ss2; ss2<<setfill('')<<setw()<<tt; ss2>>ss; for(i=;i<ss.length();i++) { a1[i]=ss[i]-''; a2[i]=a1[i]; } sort(a1,a1+ss.length(),cmp1); sort(a2,a2+ss.length(),cmp2); c1=; c2=; for(i=;i<ss.length();i++) { c1=c1*+a1[i]; c2=c2*+a2[i]; } if(c1-c2==tt&&!fir) break; else { fir=false; cout<<setfill('')<<setw()<<c1<<" - "<<setfill('')<<setw()<<c2<<" = "<<setfill('')<<setw()<<c1-c2<<endl; tt=c1-c2; } } } return ; }