Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 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.
题意:任意给出一个不超过1000位的数,让你判断它是否是一个延迟的回文数 ,设这个数字a,逆置其各位数字之后的数为b,令sum=a+b,如果在10轮计算内sum为回文数称这个数为延迟的回文数,否则就不是。
思路:写一个判断函数judge()判断给定的字符串是否为回文数,再写一个add()函数,执行字符串加法,然后模拟计算10次,观察结果。
参考代码:
#include<cstdio>
#include<cmath>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
bool judge(string s){
int n=s.size();
for(int i=0;i<n/2;i++)
if(s[i]!=s[n-i-1])
return false;
return true;
}
string add(string a,string b){
string ans="";
reverse(a.begin(),a.end());
reverse(b.begin(),b.end());
int carry=0;
for(int i=0;i<a.size();i++){
int t=a[i]-'0'+b[i]-'0'+carry;
carry=t/10;
ans+=t%10+'0';
}
if(carry) ans+=carry+'0';
reverse(ans.begin(),ans.end());
return ans;
}
int main()
{
int cnt=0,flag=false;
string a,b,sum;
cin>>a;
while(cnt<10&&!judge(a)){
b=a;
reverse(b.begin(),b.end());
sum=add(a,b);
cout<<a<<" + "<<b<<" = "<<sum<<endl;
a=sum;
cnt++;
}
if(judge(a)) cout<<a<<" is a palindromic number.\n";
else cout<<"Not found in 10 iterations.\n";
return 0;
}