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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 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:

97152

Sample Output 1:

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

Sample Input 2:

196

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.

验证一个数加上逆转它之后得到的数是否为一个回文数。 输入的数值不超过1000位，所以不可以用int或者long存储，而必须用string存储。 算法的思路很常规，将输入的数值reverse并加上数值本身，并判断它是否是回文数即可。见代码

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#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool isPalindrome(string s) {
    string t = s;
    reverse(t.begin(), t.end());
    return t == s;
}

string add(string s, string t) {
    int carry = 0;
    string result;
    for (int i = 0; i < s.length(); i++) {
        result += (s[i] + t[i] - '0' - '0' + carry) % 10 + '0';
        carry = (s[i] + t[i] - '0' - '0' + carry) / 10;
    }
    if (carry != 0) result += carry + '0';
    reverse(result.begin(), result.end());
    return result;
}

int main() {
    string s, t, sum;
    cin >> s;
    int cnt = 0;
    while (!isPalindrome(s) && cnt < 10) {
        cnt++;
        t = s;
        reverse(t.begin(), t.end());
        sum = add(s, t);
        printf("%s + %s = %s\n", s.c_str(), t.c_str(), sum.c_str());
        s = sum;
    }
    if (cnt == 10) {
        printf("Not found in 10 iterations.\n");
    } else {
        printf("%s is a palindromic number.\n", s.c_str());
    }
    return 0;
}

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下面是微博上流传的一张照片：“各位亲爱的同学们，鉴于大家有时需要使用wifi，又怕耽误亲们的学习，现将wifi密码设置为下列数学题答案：A-1；B-2；C-3；D-4；请同学们自己作答，每两日一换。谢谢合作！！~”—— 老师们为了促进学生学习也是拼了…… 本题就要求你写程序把一系列题目的答案按照卷子上给出的对应关系翻译成wifi的密码。这里简单假设每道选择题都有4个选项，有且只有1个正确答案。

输入格式：

输入第一行给出一个正整数N（<= 100），随后N行，每行按照“编号-答案”的格式给出一道题的4个选项，“T”表示正确选项，“F”表示错误选项。选项间用空格分隔。

输出格式：

在一行中输出wifi密码。

输入样例：

8
A-T B-F C-F D-F
C-T B-F A-F D-F
A-F D-F C-F B-T
B-T A-F C-F D-F
B-F D-T A-F C-F
A-T C-F B-F D-F
D-T B-F C-F A-F
C-T A-F B-F D-F

输出样例：

13224143

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#include <string>
#include <iostream>

using namespace std;

int main() {
    string a;
    cin >> a;
    while (cin >> a) {
        if (a[2] == 'T') printf("%d", a[0] - 'A' + 1);
    }
    return 0;
}
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