A1136 A Delayed Palindrome (20 point(s))

1000位的数用int，long已经不能承载，用字符串string/char[1000]

1. 原文

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.

2. 解析

反转字符串reverse() 在algorithm中

大整数之间的计算

 string temp;
int carry=0;
for (int i = str.length()-1; i >= 0 ; i--)
{
	int sum=(str[i]-'0')+(ans[i]-'0')+carry;
	temp+=sum%10+'0';
	carry=sum/10;		
}
if (carry>0)
{
	temp+=carry+'0';
}

3. AC代码

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
	string str,ans;
	cin>>str;
	int cnt=0;
    while(cnt<10){
    	ans=str;
    	reverse(str.begin(),str.end());
    	if (ans==str)
    	{
    		cout<<ans<<" is a palindromic number."<<endl;
    		return 0;
    	}
    	string temp;
    	int carry=0;
    	for (int i = str.length()-1; i >= 0 ; i--)
    	{
    		int sum=(str[i]-'0')+(ans[i]-'0')+carry;
    		temp+=sum%10+'0';
    		carry=sum/10;		
    	}
    	if (carry>0)
    	{
    		temp+=carry+'0';
    	}
    	reverse(temp.begin(),temp.end());
    	cout<<ans<<" + "<<str<<" = "<<temp<<endl;
    	str=temp;
    	cnt++;
    }
    if (cnt==10)
    {
    	cout<<"Not found in 10 iterations."<<endl;
    }
	return 0;
}