PAT 1136 A Delayed Palindrome [字符串反转] [大整数运算]

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

--------------------------------------这是题目和解题的分割线--------------------------------------

我怎么感觉这道题之前好像出现过（思考.jpg）

题目意思很简单啦，就是反转再相加，看能不能是回文。字符串反转的方法这道题总结过 PAT 1077 Kuchiguse ，这次用的是string里的reverse函数。有个需要注意的地方，不能直接相加，不然最后一个测试点会运行时错误。题目说了是十位数字，得用大整数，知识点见【大整数运算】大整数的存储 | 高精度加法 | 高精度减法。还有就是，顺序很重要！Ａ+Ｂ，A在前还是B在前很重要，不然错了测试点都不知道怎么回事！

还有两个小发现。

① 关于 no match for 'operator+' (operand types are 'basic_string' and 'int') 的报错。在解题的过程中，我发现c = (index + '0') + c; 会出现这个错误，而改写成 c += index + '0'; 则不会。查了下，看到这篇博客的解释是，编译器把括号里的内容变成了int类型（‘0’表示的是48），必须强制类型转换一下→c = char(index + '0') + c;

② string类型，如果要s[0]、s[1]......这样去赋值，必须事先给该变量一个初始值，不然赋值无效。比如说，必须string s = a;再s[0] = 'a';s[1] = 'b'。而不能string s;再s[0] = 'a';s[1] = 'b'。（我经常在string上踩雷= =以后得多用用才能多发现问题啊！

#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>using namespace std;//大整数相加 
string addBig(string a,string b)
{//这里必须有初始值 string c = a;int index = 0,i;//a、b长度一样，无需max()，也可以直接倒着遍历 for(i=a.length()-1;i>=0;i--){//别忘了数字和字符之间的相互转换('0') int tmp = a[i] -'0' + b[i] -'0' + index;c[i] = tmp%10 + '0';index = tmp/10;}//如果还有进位，加到字符串的开头 if(index!=0)c = char(index + '0') + c;return c;
}int main()
{string str,newStr;cin>>str;newStr = str;int cnt = 0;reverse(str.begin(),str.end());//循环条件，不是回文并且次数在10以内 while(newStr!=str&&cnt<10){cout<<newStr<<" + "<<str<<" = ";newStr = addBig(str,newStr);str = newStr;cout<<newStr<<endl;reverse(str.begin(),str.end());cnt++;}if(cnt>=10)cout<<"Not found in 10 iterations.\n";elsecout<<newStr<<" is a palindromic number.\n";return 0;
}

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