PAT Grade 1024 Palindromic Number (25 minutes) (addition of large numbers, consider this number is not a palindrome sequence beginning)

1024 Palindromic Number (25 分)

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

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. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

answer:

　　CONSIDERING the outset with the addition of large numbers just in case. First submission found 2 test points and test points 3 have not seen their own analysis of the reasons may have been before the string is a palindrome without, that is, k = 0, there is a single digit palindrome string.

5 100
5
0

131 500
131

　　This title examines a number of addition and reverse, this belongs to the simple question, but the nature of the test Tarsus sum of knowledge. Not adding lead to large numbers noted test points 6 and 8 test points (starting from 0) is not adopted. The reason is this question is N range (0,1010], the range of k is (0, 100], we consider the worst case, assuming that N is a very approximate value of 1010 and 100 were still poorly step operation value palindrome , is simply the maximum calculated estimation understood encountered is 2100 * 1010, the value exceeds the long long int (263-1, approximately 9.2 * 1018) indicates a range, it is necessary char value stored number.

AC Code:

#include <bits / STDC ++ H.>
 the using  namespace STD;
 char A [ 101 ];
 char B [ 101 ];
 char C [ 101 ];
 int n-;
 int main () { 
    CIN >> A; 
    CIN >> n-;
     int = L strlen (A);
     for ( int I = 0 ; I <L; I ++ ) { 
        B [I] = A [Li- . 1 ]; 
    } 
    // check generation 0 is not itself palindromic 
    int F = 1 ; 
     int mid=(l-1)/2;
    for(int j=0;j<=mid;j++){
        if(a[j]!=a[l-j-1]){
            f=0;
            break;
        }
    }
    if(f){
        cout<<a<<endl;
        cout<<0;
    }
    else{//再考虑第1代及以后 
        int k=-1;
        for(int i=1;i<=n;i++) {
             // add up a new value 
            int X = 0 ;
             for ( int J = 0 ; J <L; J ++ ) { 
                X = A [J] - ' 0 ' + B [J] - ' 0 ' + X; 
                C [J] = X% 10 + ' 0 ' ; 
                X = X / 10 ; 
            } 
            IF (X> 0 ) { 
                C [L ++] = X + ' 0 '  ;
            }
             // check the character does not meet the requirements 
            MID = (L- . 1 ) / 2 ; 
            F = . 1 ;
             for ( int J = 0 ; J <= MID; J ++ ) {
                 IF ! (C [J] = C [lj- . 1 ]) { 
                    F = 0 ;
                     BREAK ; 
                } 
            } 
            IF (F) { // palindromic 
                K = I;
                 BREAK ; 
            } 
            for ( int= J endl;0 ; J <L; J ++) { // update 
                A [J] = C [J]; 
                B [J] = C [lj- . 1 ]; 
            }          
        } 
        for ( int I = L- . 1 ; I> = 0 ; I -) { // output 
            COUT << C [I]; 
        } 
        COUT << endl;
         IF ! (K = - . 1 ) { // not to n generations 
            COUT << K << endl; 
        } the else {      
            COUT << << n- 
        }    
    }
    return 0;
}

Learn about other people too concise Code:

#include <iostream>
#include <algorithm>
using namespace std;
string add(string a){
    string ans=a;
    reverse(a.begin(),a.end());
    int i=a.length()-1,add=0;
    while(i>=0){
        int tmp=a[i]-'0'+ans[i]-'0';
        ans[i]=(add+tmp)%10+'0';
        add=(tmp+add)/10;
        i--;
    }
    if(add) ans.insert(0,"1");
    return ans;
}
int main(){
    string s;
    int k;
    cin>>s>>k;
    string tmp=s;
    reverse(tmp.begin(),tmp.end());
    if(tmp==s) cout<<tmp<<endl<<0;
    else{
        int i=0;
        while(i<k){
            s=add(tmp);
            i++;
            tmp=s;
            reverse(tmp.begin(),tmp.end());
            if(tmp==s) break;
        }
        cout<<s<<endl<<i;
    } 
    return 0;
}