• PAT 1069 The Black Hole of Numbers


    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 (.

    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

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    
    bool comp(char a,char b){
        return a>b;
    }
    
    int to_int(string s){
        int sum = 0;
        for(int i=0;i < s.size();i++){
            int num = s[i] - '0';
            sum = sum*10 + num;
        }
        return sum;
    }
    
    
    string minuss(string s1,string s2){
        string res;
        int a = to_int(s1);
        int b = to_int(s2);
        int c = a - b;
        res = to_string(c);
        int len = res.size();
        for(int i=0;i < (4-len);i++){ //加前导0;
            res = "0"+res;
        }
        return res;
    }
    
    int main(){
    
        string s;cin >> s;
        int len = s.size();
        for(int i=0;i < (4-len);i++){ //加前导0;
            s = "0"+s;
        }
        string s1 = s,s2 = s;
        sort(s1.begin(),s1.end(),comp);
        sort(s2.begin(),s2.end());
        string ans = minuss(s1,s2);
    //    cout << ans;
        if(ans == "0000"){
            printf("%s - %s = %s
    ",s1.c_str(),s2.c_str(),ans.c_str());
            return 0;
        }
        else{
            printf("%s - %s = %s
    ",s1.c_str(),s2.c_str(),ans.c_str());
        }
    
        while(ans!="6174"){
            s1 = ans; s2 = ans;
            sort(s1.begin(),s1.end(),comp);
            sort(s2.begin(),s2.end());
            ans = minuss(s1,s2);
            printf("%s - %s = %s
    ",s1.c_str(),s2.c_str(),ans.c_str());
        }
    
        return 0;
    }

    贼弱智,说是四位数,你给个0~10000范围,补前导0还算数字,服了

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  • 原文地址:https://www.cnblogs.com/cunyusup/p/10772040.html
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