1136 A Delayed Palindrome (20 分)
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.
题目大意:判断回文数字,如果不是回文数字就经过一系列计算并输出计算过程。需要注意A和B的位数相同,即B的高位需要补零。数字的位数不超过1000,所以需要用字符串存储数字并另外写个函数进行逐位相加的操作。
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
bool isPalindrome(string &s);
string Sum(const string &A,const string &B);
int main()
{
int cnt=0;
string A,B,sum;
cin>>sum;
if(isPalindrome(sum)){
printf("%s is a palindromic number.\n",sum.c_str());//如果是回文的话直接输出
return 0;
}
while(1){
A=sum;
B.resize(A.size());
reverse_copy(A.begin(),A.end(),B.begin());
sum=Sum(A,B);
cout<<A<<" + "<<B<<" = "<<sum<<endl;
cnt++;
if(isPalindrome(sum)){
cout<<sum;
printf(" is a palindromic number.\n");
break;
}
if(cnt==10){
printf("Not found in 10 iterations.\n");
break;
}
}
return 0;
}
string Sum(const string &A,const string &B)
{
string sum;
char c,next='0';
for(int i=A.size()-1;i>=0;i--){
c=A[i]+B[i]+next-'0'-'0';
if(c<='9'){
sum.insert(sum.begin(),c);
next='0';
}
else{//当前位的数字相加大于10,则高位+1,当前位-10
sum.insert(sum.begin(),c-'9'+'0'-1);
next='1';
}
}
if(next!='0') sum.insert(sum.begin(),next);
return sum;
}
bool isPalindrome(string &s)
{
if(s.size()==1) return true;
if(s.size()==2){
if(s[1]!=s[0]&&s[1]!=0) return false;
else return true;
}
int k=s.size();
if(s[0]!=s[k-1]&&s[k-1]!=0) return false;
for(int i=1,j=k-2;i<j;i++,j--)
if(s[i]!=s[j]) return false;
return true;
}