1136 A Delayed Palindrome (20 分)
标签：numbers min lis test end using ecif wing code
Consider a positive integer N written in standard notation with k+1 digits a?i?? as a?k???a?1??a?0?? with 0 for all iand 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.
Each input file contains one test case which gives a positive integer no more than 1000 digits.
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.
97152
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
196
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.
1 #include <bits/stdc++.h> 2 #define ll long long int 3 using namespace std; 4 string s,st; 5 6 7 string add(string s, string ss){ 8 string st; 9 int an = 0; 10 for(int i = s.length()-1 ; i >= 0; --i){ 11 int x = s[i]-‘0‘; 12 int y = ss[i]-‘0‘; 13 int ans = x+y+an; 14 if(ans < 10){ 15 st += ans+‘0‘; 16 an = 0; 17 }else{ 18 st += (ans%10) + ‘0‘; 19 an = 1; 20 } 21 } 22 if(an) st += ‘1‘; 23 reverse(st.begin(), st.end()); 24 return st; 25 } 26 27 28 bool ishui(string s){ 29 for(int i = 0; i < s.length(); i++){ 30 if(s[i] != s[s.length() -i - 1]) 31 return false; 32 } 33 return true; 34 } 35 36 int main(){ 37 cin >> s; 38 st = s; 39 reverse(st.begin(), st.end()); 40 int cnt = 10; 41 while(!ishui(s) && cnt--){ 42 string ans = add(s,st); 43 cout <<s<<" + "<<st<<" = "<<ans<<endl; 44 s = ans; 45 st = s; 46 reverse(st.begin(), st.end()); 47 } 48 if(cnt == -1){ 49 cout <<"Not found in 10 iterations."<<endl; 50 }else{ 51 cout << s<<" is a palindromic number."<<endl; 52 } 53 return 0; 54 }
1136 A Delayed Palindrome (20 分)
标签：numbers min lis test end using ecif wing code
原文地址：https://www.cnblogs.com/zllwxm123/p/11296164.html
