PAT 甲级 1136 A Delayed Palindrome (20分)
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.
【AC】还差最后一个点、查了半个多小时、要1点了、不想找了、暂时就这样吧、想到再改
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 string s, s2; cin >> s; s2 = s; 7 reverse(s2.begin(), s2.end()); 8 if(s == s2) { cout << s << " is a palindromic number."; return 0; } 9 int cnt = 0; string s1 = s; 10 while(1) 11 { 12 int flag1 = 1; 13 reverse(s1.begin(), s1.end()); 14 int sum = stoi(s) + stoi(s1); 15 string ss = to_string(sum); 16 for(int i = 0; i < ss.length()/2; i++) 17 { 18 if(ss[i] != ss[ss.length()-i-1]) { flag1 = 0; break; } 19 } 20 if(flag1) { cout << s << " + " << s1 << " = " << ss << endl << ss << " is a palindromic number."; return 0; } 21 cnt++; cout << s << " + " << s1 << " = " << ss << endl; 22 s = s1 = ss; 23 if(cnt == 10) { cout << "Not found in 10 iterations."; return 0; } 24 } 25 return 0; 26 }
【柳婼AC代码】
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 string rev(string s) { 5 reverse(s.begin(), s.end()); 6 return s; 7 } 8 string add(string s1, string s2) { 9 string s = s1; 10 int carry = 0; 11 for (int i = s1.size() - 1; i >= 0; i--) { 12 s[i] = (s1[i] - '0' + s2[i] - '0' + carry) % 10 + '0'; 13 carry = (s1[i] - '0' + s2[i] - '0' + carry) / 10; 14 } 15 if (carry > 0) s = "1" + s; 16 return s; 17 } 18 int main() { 19 string s, sum; 20 int n = 10; 21 cin >> s; 22 if (s == rev(s)) { 23 cout << s << " is a palindromic number.\n"; 24 return 0; 25 } 26 while (n--) { 27 sum = add(s, rev(s)); 28 cout << s << " + " << rev(s) << " = " << sum << endl; 29 if (sum == rev(sum)) { 30 cout << sum << " is a palindromic number.\n"; 31 return 0; 32 } 33 s = sum; 34 } 35 cout << "Not found in 10 iterations.\n"; 36 return 0; 37 }
Tips:reverse( str.begin(), str.end() ); //palindrome重要函数 反转字符串