The Black Hole of Numbers (strtoint+inttostr+sort)

 

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

 

坑点1、数字都要四位的2、如果是判断下一个式子的差不等于上个结果，然后跳出的话，那么需要判断，是不是第一次输出。比如，输入6174，结果就等于6174，那么就会没输出，直接跳出。

 

  1 #include <iostream>

  2 

  3 #include <algorithm>

  4 

  5 #include<string>

  6 

  7 #include <sstream>

  8 

  9 #include <iomanip>

 10 

 11 using namespace std;

 12 

 13  

 14 

 15 int a1[5];

 16 

 17 int a2[5];

 18 

 19  

 20 

 21 int bb[1000];

 22 

 23  

 24 

 25 bool cmp1(int a,int b)

 26 

 27 {

 28 

 29    return a>b;

 30 

 31 }

 32 

 33  

 34 

 35 bool cmp2(int a,int b)

 36 

 37 {

 38 

 39    return a<b;

 40 

 41 }

 42 

 43  

 44 

 45 int main()

 46 

 47 {

 48 

 49       string n;

 50 

 51       int i;

 52 

 53  

 54 

 55       while(cin>>n)

 56 

 57       {

 58 

 59       

 60 

 61          int tt,c1,c2;

 62 

 63          stringstream ss1;

 64 

 65          ss1<<n;

 66 

 67          ss1>>tt;

 68 

 69       

 70 

 71          i=0;

 72 

 73          bool fir=true;

 74 

 75          while(true)

 76 

 77          {

 78 

 79  

 80 

 81                   string ss;

 82 

 83                   stringstream ss2;

 84 

 85                   ss2<<setfill('0')<<setw(4)<<tt;

 86 

 87                   ss2>>ss;

 88 

 89  

 90 

 91                   for(i=0;i<ss.length();i++)

 92 

 93                   {

 94 

 95                  a1[i]=ss[i]-'0';

 96 

 97                    a2[i]=a1[i];

 98 

 99                   }

100 

101  

102 

103              sort(a1,a1+ss.length(),cmp1);

104 

105                sort(a2,a2+ss.length(),cmp2);

106 

107  

108 

109                    c1=0; c2=0;

110 

111                for(i=0;i<ss.length();i++)

112 

113                   {

114 

115                    c1=c1*10+a1[i];

116 

117                      c2=c2*10+a2[i];

118 

119                   }

120 

121  

122 

123                   

124 

125                   if(c1-c2==tt&&!fir)  break;

126 

127                   else 

128 

129                         

130 

131                   {

132 

133                         fir=false;

134 

135                         cout<<setfill('0')<<setw(4)<<c1<<" - "<<setfill('0')<<setw(4)<<c2<<" = "<<setfill('0')<<setw(4)<<c1-c2<<endl;

136 

137                         tt=c1-c2;

138 

139                   }     

140 

141          }  

142 

143       }

144 

145    return 0;

146 

147 }

148 

149  

View Code