1069. The Black Hole of Numbers (20)

本文介绍了一个有趣的数学现象——通过特定操作使得任意一个不含完全相同数字的四位数最终都会到达一个固定的数值6174,即Kaprekar常数。并提供了一段C++代码实现这一过程。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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

用string 来接受参数比较方便

参考代码:

#include <iostream>
#include <algorithm>
using namespace std;
int compare(char a,char b){
    return a>b;
}
string getStandard(string s){
    if(s.size()<4){
        int x = 4-s.size();
        for(int i=0;i<x;i++){
            s = "0"+s;
        }
    }
    char *temp = new char[s.size()];
    for(int i=0;i<s.size();i++){
        temp[i] = s[i];
    }
    sort(temp,temp+s.size(),compare);
    string result = "";
    for(int i=0;i<s.size();i++){
        result += temp[i];
    }
    return result;
}
string reverseIt(string s){
    string temp = "";
    for(int i=0;i<s.size();i++){
        temp += s[s.size()-1-i];
    }
    return temp;
}
string caculate(string s1,string s2){
    int mark = 0;
    string result = "";
    for(int i=s1.size()-1;i>=0;i--){
        if(mark==0){
            if(s2[i]>=s1[i]){
                char ch = s2[i]-s1[i]+'0';
                result = ch+result;
                mark = 0;
            }else{
                char ch = s2[i]-s1[i]+10+'0';
                result = ch+result;
                mark = 1;
            }
        }else{
            if(s2[i]>s1[i]){
                char ch = s2[i]-s1[i]-1+'0';
                result = ch +result;
                mark = 0;
            }else{
                char ch = s2[i]-s1[i]+10+'0'-1;
                result = ch+result;
                mark = 1;
            }
        }
    }
    return result;
}
string N;
int main()
{
    cin>>N;
    string standard = getStandard(N);
    if(standard==reverseIt(standard)){
        cout<<standard<<" - "<<standard<<" = "<<"0000";
        return 0;
    }
    while(standard!="6174"){
        string reverseS = reverseIt(standard);
        string result = caculate(reverseS,standard);
        cout<<standard<<" - "<<reverseS<<" = "<<result<<endl;
        if(result=="6174"){
            break;
        }
        standard = getStandard(result);
    }
    return 0;
}


 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值