• PAT 1069 The Black Hole of Numbers[简单]


    1069 The Black Hole of Numbers(20 分)

    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 (0,104​​).

    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

     题目大意:给定一个四位数,展示出每位数从大到小排列,与从小到大排列的差值,直到出现6174黑洞数停止。

    //这道题目是简单的,但是还是遇见了一些问题:

    AC:

    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <stdlib.h>
    #include<cstdio>
    using namespace std;
    
    int main()
    {
        int n;
        cin>>n;
        int a[4],big,small,res=-1;
        int temp=0;
        while(res!=6174)
        {
            fill(a,a+4,0);
            while(n!=0)
            {
                a[temp++]=n%10;
                n/=10;
                //cout<<"kk";
            }
            temp=0;
            sort(a,a+4);//默认从小到大排列
            small=a[0]*1000+a[1]*100+a[2]*10+a[3];
            big=a[3]*1000+a[2]*100+a[1]*10+a[0];
            if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3])
            {
                printf("%04d - %04d = 0000
    ",big,small);
                return 0;
            }
            res=big-small;
            printf("%04d - %04d = %04d
    ",big,small,res);
            n=res;
        }
        return 0;
    }

    1.第一次忽略了temp=0;应该在使用过后将其赋值为0的;

    2. 应该将a数组初始化为0,要不然下次会有影响的,比如在有0位的时候:

    3.最后就是提交发现第0个测试点过不去,原来是因为相同数的时候只输出了0,而不是0000!

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  • 原文地址:https://www.cnblogs.com/BlueBlueSea/p/9584341.html
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