文章目录
Author: CHEN, Yue
Organization: 浙江大学
Time Limit: 400 ms
Memory Limit: 64 MB
Code Size Limit: 16 KB
A1136 A Delayed Palindrome (20point(s))
Consider a positive integer N written in standard notation with k+1 digits ai as ak ⋯a1 a0 with 0≤ai <10 for all i and ak >0. Then N is palindromic if and only if ai =ak−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.
Code
#include <bits/stdc++.h>
using namespace std;
bool isPal(string s){
for(int i=0;i<s.size()/2;i++){
if(s[i]!=s[s.size()-i-1])
return false;
}
return true;
}
string stringSum(string a,string b){
int res[1010];
string s;
memset(res,-1,sizeof(res));
for(int i=0;i<a.size();i++) res[i]=(a[i]-'0'+b[i]-'0');
int i=0;
while(res[i]!=-1){
if(res[i]>=10){
if(res[i+1]==-1) res[i+1]=0;
res[i+1]+=res[i]/10;
res[i]%=10;
}
char t=res[i++]+'0';
s+=t;
}
reverse(s.begin(),s.end());
return s;
}
int main(){
int n;
string s,rs;
cin>>s;
if(isPal(s)==true) cout<<s<<" is a palindromic number.\n";
else{
for(int i=0;i<10;i++){
rs=s;
reverse(rs.begin(),rs.end());
if(isPal(s)==true){
cout<<s<<" is a palindromic number.\n";
return 0;
}
else cout<<s<<" + "<<rs<<" = "<<stringSum(s,rs)<<"\n";
s=stringSum(s,rs);
}
cout<<"Not found in 10 iterations.\n";
}
return 0;
}
Analysis
-给出一个初始的数,判断是否为回文数。若不是就将其与其每位逆转后的数相加,若仍然不是,则重复操作。
-若进行了10次相加操作,仍然得不到回文数,则输出"Not found in 10 iterations."。