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 }