PAT 1136 A Delayed Palindrome（高精度加法）

1136 A Delayed Palindrome

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.

总结： 高精度加法，不过最后有一个坑点，后面的英文句子最后不能加上回车符！

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

vector<int> add(vector<int> & a,vector<int> &b){
    vector<int> c;
    int t=0;
    
    for(int i=0;i<a.size();i++){
        t=t+a[i]+b[i];
        c.push_back(t%10);
        t=t/10;
    }
    if(t>0) c.push_back(1);
    return c;
}

bool judge(vector<int> &a,vector<int> &b){
    for(int i=0;i<a.size();i++)
        if(a[i]!=b[i])  return false;
    return true;
}

int main(){
    string s;
    int cot=0;
    vector<int> a,b;
    cin >> s;
    for(int i=0;i<s.size();i++) a.push_back(s[i]-'0');
    
    while(cot<10){
        vector<int> b=a;
        reverse(a.begin(),a.end());
        if(judge(a,b))  break;
        auto c=add(a,b);
        for(int i=0;i<b.size();i++) printf("%d",b[i]);
        printf(" + ");
        for(int i=0;i<a.size();i++) printf("%d",a[i]);
        printf(" = ");
        for(int i=c.size()-1;i>=0;i--) printf("%d",c[i]);
        reverse(c.begin(),c.end());
        a=c;
        cot++;
        puts("");
    }
    if(cot<10){
        for(int i=0;i<a.size();i++) printf("%d",a[i]);
        printf(" is a palindromic number.");
    }
    else puts("Not found in 10 iterations.");
    
    return 0;
}

柳神代码：

#include <iostream>
#include <algorithm>
using namespace std;

string add(string a){
	string b=a,ans;
	reverse(a.begin(),a.end());
	int carry=0;
	for(int i=a.size()-1;i>=0;i--){
		int num=(a[i]-'0'+b[i]-'0')+carry;
		carry=0;
		if(num>=10){
			num%=10;
			carry=1;
		}
		ans+=char(num+'0');//因为是先从个位进行计算的，所以最先存下来的是个位
	}
	if(carry==1)	ans+=char(1+'0');
    reverse(ans.begin(),ans.end());
	return ans;
}
int main(){
	string s;
	cin >> s;
	int cot=0;
    
	while(cot<10){
		string t=s;
		reverse(s.begin(),s.end());
		if(s==t){
			cout << s << " is a palindromic number.";
			break;
		}
		else{
			cout << t << " + " << s << " = " << add(t) << endl;
			s=add(t);
			cot++;
		}
	}
	if(cot==10)	cout << "Not found in 10 iterations.";
	return 0;
}

好好学习，天天向上！

我要考研！