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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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原文地址：http://www.cnblogs.com/xiaoyesoso/p/4240612.html
