PAT甲级:1136 A Delayed Palindrome (20分)
题干
Consider a positive integer N written in standard notation with k+1 digits a**i as a**k⋯a1a0 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
实际上很简单,给大家提一下坑点:
- 首次输入的数字可能已经符合条件,不需要再作运算。例如11就可以直接输出答案,不需要再运算1次.
- 输入的数字用
long long int
都没有办法全部存上,必须用string
实现模拟加法。
code
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string palindromic(string num, string r_num){
string ans;
int flag = 0;
for(int i = 0; i < num.size(); i++){
int temp = num[i] + r_num[i] - 2 * '0' + flag;
flag = temp / 10;
ans.push_back((char) temp % 10 + '0');
}
if(flag) ans.push_back(flag + '0');
reverse(ans.begin(), ans.end());
printf("%s + %s = %s
", num.c_str(), r_num.c_str(), ans.c_str());
return ans;
}
int main(){
string num;
cin >> num;
for(int i = 0; i < 10; i++){
string r_temp = num;
reverse(r_temp.begin(), r_temp.end());
if(num == r_temp){
printf("%s is a palindromic number.", num.c_str());
return 0;
}
num = palindromic(num, r_temp);
}
printf("Not found in 10 iterations.");
return 0;
}