A1069. The Black Hole of Numbers

时间:2022-09-26 17:32:50

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
 #include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp1(int a, int b){
return a < b;
}
bool cmp2(int a, int b){
return a > b;
}
void numSort(int n, int &r1, int &r2){
int temp[];
int i = ;
r1 = ; r2 = ;
do{
temp[i++] = n % ;
n = n / ;
}while(n != || i < );
sort(temp, temp + i, cmp1);
for(int j = , P = ; j < i; j++){
r1 = r1 + P * temp[j];
P = P * ;
}
sort(temp, temp + i, cmp2);
for(int j = , P = ; j < i; j++){
r2 = r2 + P * temp[j];
P = P * ;
}
}
int main(){
int N, r1, r2, ans;
scanf("%d", &N);
numSort(N, r1, r2);
do{
ans = r1 - r2;
printf("%04d - %04d = %04d\n", r1, r2, ans);
numSort(ans, r1, r2);
}while(ans != && ans != );
cin >> N;
return ;
}

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)。