PAT A1136 A Delayed Palindrome （20 分）

2019-02-23 21:42:43 阅读：242 来源： 互联网

标签：A1136 Palindrome PAT string palindromic s2 number res include

Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0≤ai<10 for all i and ak>0. Then N is palindromic if and only if ai=ak−i for all i. Zero is written 0 and is also palindromic by definition.

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. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

Input Specification:

Each input file contains one test case which gives a positive integer no more than 1000 digits.

Output Specification:

For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

A + B = C

where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations.instead.

Sample Input 1:

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

Sample Output 2:

196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

 1 #include <stdio.h>
 2 #include <string>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <string.h>
 7 using namespace std;
 8 string add(string s1,string s2){
 9     string res="";
10     int carry=0;
11     for(int i=s1.length()-1;i>=0;i--){
12         int tmp=s2[i]-'0'+s1[i]-'0'+carry;
13         res+=(tmp%10+'0');
14         carry=tmp/10;
15     }
16     if(carry!=0)res += (carry+'0');
17     reverse(res.begin(),res.end());
18     return res;
19 }
20 bool ispali(string s){
21     string s2;
22     s2=s;
23     reverse(s2.begin(),s2.end());
24     if(s==s2) return true;
25     else return false;
26 }
27 int main(){
28       string s1,s2,s3;
29       cin>>s1;
30       int i;
31       if(ispali(s1)){
32           cout<<s1<<" is a palindromic number."<<endl;
33               return 0;
34       }
35       for(i=0;i<10;i++){
36           s2=s1;
37           reverse(s2.begin(),s2.end());
38           s3=add(s1,s2);
39           cout<<s1<<" + "<<s2<<" = "<<s3<<endl;
40           if(ispali(s3)){
41               cout<<s3<<" is a palindromic number."<<endl;
42               return 0;
43           }
44           s1=s3;
45       }
46       printf("Not found in 10 iterations.\n");
47 }

View Code

注意点：判断回文用string还是方便，string只能string+char，不能char+string，所以大数相加还是要最后反转一下。

标签：A1136,Palindrome,PAT,string,palindromic,s2,number,res,include
来源： https://www.cnblogs.com/tccbj/p/10424328.html

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