题目信息

1069. The Black Hole of Numbers (20)

时间限制100 ms

内存限制65536 kB

代码长度限制16000 B

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

解题思路

直接计算就可以

AC代码

#include <cstdio>

#include <algorithm>

using namespace std;

int getmin(int a){

    int t[4] = {0}, c = 0;

    while (a){

        t[c++] = a%10;

        a /= 10;

    }

    sort(t, t + 4);

    c = 0;

    while (c < 4){

        a = a*10 + t[c++];

    }

    return a;

}

int getmax(int a){

    int t[4] = {0}, c = 0;

    while (a){

        t[c++] = a%10;

        a /= 10;

    }

    sort(t, t + 4, greater<int>());

    c = 0;

    while (c < 4){

        a = a*10 + t[c++];

    }

    return a;

}

int main()

{

    int a;

    scanf("%d", &a);

    while (true){

        int mn = getmin(a);

        int mx = getmax(a);

        printf("%04d - %04d = %04d\n", mx, mn, mx - mn);

        a = mx - mn;

        if (a == 6174 || mn%10 == mn/1000) break;

    }

    return 0;

}

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