• 1069. The Black Hole of Numbers (20)


    题目如下:

    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
    


    题目要求将一个四位数按照降序、升序生成减数和被减数然后作差,如果差等于0或者6174则停止运算。

    题目的注意点在于数字不够4位时要前补0。

    为了方便的排序四位数,我们先使用vector容纳四位然后排序,接着利用stringstream将vecotr中元素转成数字,因为输出时减数、被减数和差都需要,因此用一个结构体存储每次的运算结果。

    代码如下:

    #include <iostream>
    #include <vector>
    #include <sstream>
    #include <algorithm>
    #include <sstream>
    #include <stdio.h>
    
    using namespace std;
    
    struct Result{
        int num1;
        int num2;
        int result;
    }res;
    
    void compute(int input){
        vector<int> num1,num2;
        num1.push_back(input / 1000);
        num1.push_back(input % 1000 / 100);
        num1.push_back(input % 100 / 10);
        num1.push_back(input % 10);
        num2 = num1;
        sort(num1.begin(),num1.end(),less_equal<int>());
        sort(num2.begin(),num2.end(),greater_equal<int>());
        stringstream ss;
        for(int i = 0; i < 4; i++){
            ss << num1[i];
        }
        int dec1;
        ss >> dec1;
        ss.clear();
        for(int i = 0; i < 4; i++){
            ss << num2[i];
        }
        int dec2;
        ss >> dec2;
        res.num1 = dec2;
        res.num2 = dec1;
        res.result = dec2 - dec1;
    }
    
    int main()
    {
        int input;
        cin >> input;
        int i = 5;
        do{
            compute(input);
            input = res.result;
            if(input == 0){
                printf("%04d - %04d = %04d
    ",res.num1,res.num2,input);
                break;
            }else{
                printf("%04d - %04d = %04d
    ",res.num1,res.num2,input);
            }
        }while(input != 6174);
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/aiwz/p/6154083.html
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