标签：contain board inner mes red one art int word
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 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.
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.这道题考察了大数的加减、回文数。
c++解法：
#include <iostream> #include <algorithm> using namespace std; string plus_str(string s1,string s2){ reverse(s1.begin(),s1.end()); reverse(s2.begin(),s2.end()); string res="";int i;bool jinwei=false; for(i=0;i<s2.size();i++){ if(jinwei) { res+=((s1[i]-‘0‘+s2[i]+1-‘0‘)%10+‘0‘); jinwei=false; if((s1[i]-‘0‘+s2[i]-‘0‘+1)/10>0) jinwei=true; } else { res+=((s1[i]-‘0‘+s2[i]-‘0‘)%10+‘0‘); if((s1[i]-‘0‘+s2[i]-‘0‘)/10>0) jinwei=true; } } if(jinwei) res+=‘1‘; reverse(res.begin(),res.end()); return res; } int main() { string str,rev,add;int n=10; cin>>str; while(n--){ rev=str; reverse(rev.begin(),rev.end()); if(str==rev) { printf("%s is a palindromic number.",str.data()); system("pause"); return 0; } add=plus_str(str,rev); printf("%s + %s = %s\n",str.data(),rev.data(),add.data()); str=add; } printf("Not found in 10 iterations.");
return 0; }
Java解法
import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); BigInteger num = sc.nextBigInteger(); for(int i = 0; i < 10; i++){ BigInteger revNum = new BigInteger(new StringBuilder(num+"").reverse().toString()); if(num.equals(revNum)) { System.out.printf("%s is a palindromic number.\n", num.toString()); return ; } else { BigInteger plus = num.add(revNum); System.out.printf("%s + %s = %s\n", num, revNum, plus); num = plus; } } System.out.println("Not found in 10 iterations."); } }
PAT Advanced 1136 A Delayed Palindrome (20分)
标签：contain board inner mes red one art int word
原文地址：https://www.cnblogs.com/littlepage/p/12236281.html
