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


    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
    ", mx, mn, mx - mn);
            a = mx - mn;
            if (a == 6174 || mn%10 == mn/1000) break;
        }
        return 0;
    }

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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/7238711.html
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