Palindromic Number (还是大数)

 

 

 

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

 

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

 

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

 

Input Specification:

 

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (<= 1010) is the initial numer and K (<= 100) is the maximum number of steps. The numbers are separated by a space.

 

 Output Specification:

 

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

 Sample Output 1:

484

2

 Sample Input 2:

69 3

 Sample Output 2:

1353

3

 

  1 #include <iostream>

  2 

  3 #include <string>

  4 

  5 #include <algorithm>

  6 

  7 using namespace std;

  8 

  9  

 10 

 11 int aa1[50];

 12 

 13 int aa2[50];

 14 

 15  

 16 

 17 int main()

 18 

 19 {

 20 

 21  

 22 

 23       string  n;int k;

 24 

 25     while(cin>>n)

 26 

 27       {

 28 

 29             cin>>k;

 30 

 31           int i,j,t;

 32 

 33  

 34 

 35        bool ifid=true;

 36 

 37  

 38 

 39          for(i=0,j=n.length()-1;i<=j;i++,j--)

 40 

 41          {

 42 

 43              if(n[i]!=n[j])

 44 

 45                {

 46 

 47                  ifid=false;

 48 

 49                   break;

 50 

 51                }

 52 

 53          }

 54 

 55  

 56 

 57          if(ifid)

 58 

 59          {

 60 

 61             cout<<n<<endl;

 62 

 63               cout<<0<<endl;

 64 

 65          }

 66 

 67          else

 68 

 69          {

 70 

 71                 for(i=0;i<50;i++)

 72 

 73                   {

 74 

 75                     aa1[i]=0;

 76 

 77                        aa2[i]=0;

 78 

 79                   }

 80 

 81                 int count=0;

 82 

 83                 for(i=n.length()-1;i>=0;i--)

 84 

 85                   {

 86 

 87                   aa1[count]=n[i]-'0';

 88 

 89                     aa2[count]=n[i]-'0';

 90 

 91                     count++;

 92 

 93                   }

 94 

 95  

 96 

 97                 reverse(aa2,aa2+count);

 98 

 99               int tem=0;

100 

101                   int sum=0;

102 

103                 for(i=1;i<=k;i++)

104 

105                   {

106 

107                      for(j=0;j<count;j++)

108 

109                               aa1[j]=aa1[j]+aa2[j];

110 

111                         sum++;

112 

113                  for(j=0;j<count;j++)

114 

115                                 {

116 

117                               if(aa1[j]>9)

118 

119                                       {

120 

121                                  tem=aa1[j]/10;

122 

123                                  aa1[j+1]=aa1[j+1]+tem;

124 

125                                  aa1[j]=aa1[j]%10; 

126 

127                                       }

128 

129                                 }

130 

131                         if(aa1[j]!=0) count++;

132 

133  

134 

135                         

136 

137                   bool ifis=true;

138 

139  

140 

141                     for(j=0,t=count-1;j<=t;j++,t--)

142 

143                           {

144 

145                          if(aa1[j]!=aa1[t])

146 

147                                  {

148 

149                              ifis=false;

150 

151                                break;

152 

153                                  }

154 

155                           }

156 

157  

158 

159                     if(ifis)

160 

161                           {

162 

163                       break;

164 

165                           }

166 

167                           else

168 

169                           {

170 

171                             for(j=0;j<count;j++)

172 

173                                     aa2[j]=aa1[j];

174 

175                               reverse(aa2,aa2+count);

176 

177                           }

178 

179                   }

180 

181  

182 

183  

184 

185                   for(j=count-1;j>=0;j--)

186 

187                         cout<<aa1[j];

188 

189                   cout<<endl;

190 

191                   cout<<sum<<endl;

192 

193          }

194 

195  

196 

197       }

198 

199       return 0;

200 

201 }

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