Consider a positive integer N written in standard notation with k+1 digits a?i?? as a?k???a?1??a?0?? with 0≤a?i??<10 for all i and a?k??>0. Then N is palindromic if and only if a?i??=a?k?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.
思路: 因为他的数字个数不超过 1000 个, 所以用字符串来保存数据;
然后判断是不是回文串, 反过来比较是否相同即可;
其中用到了 reverse() 函数, 把一个字符串反过来, 大数相加, 存进位; 因为是加法, 所以进位最大也只可能是 1;
注意事项:
将单个字符存入字符串: ans += char(num + '0');
将个位数转换为 char 类型, 要 + '0'; 不能 - '0'
代码如下:
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- string add(string a)
- {
- string ans;
- int l = a.length();
- int co = 0;
- for(int i = 0;i <l;i++)
- {
- int num = (a[i] - '0') + (a[l - 1 - i] - '0') + co;
- //cout << num << "---" << endl;
- co = 0;
- if(num> 9)
- {
- co = 1;
- num -= 10;
- }
- ans += char(num + '0');
- }
- if(co == 1)
- ans += '1';
- reverse(ans.begin(),ans.end());
- //cout <<ans << endl;
- return ans;
- }
- string a,b;
- int main()
- {
- cin>> a;
- int i,len;
- for(i = 0;i < 10;i++)
- {
- b = a;
- reverse(b.begin(),b.end());
- if(a == b)
- break;
- cout << a << "+" << b << "=" << add(a) << endl;
- a = add(a);
- }
- if(i == 10)
- cout << "Not found in 10 iterations." << endl;
- else
- cout << a << "is a palindromic number." << endl;
- return 0;
- }
来源: http://www.bubuko.com/infodetail-2877364.html