題目信息

1069. The Black Hole of Numbers (20)

時間限制100 ms
內存限制65536 kB
代碼長度限制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:

解題思路

直接計算就可以

AC代碼

#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;
}

個人遊戲推廣：
《10雲方》與方塊來次消除大戰!

1069. The Black Hole of Numbers (20)【模擬】——PAT (Advanced Level) Practise