1024 Palindromic Number （25 分)

2019-03-13

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 N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (≤1010) is the initial numer and K (≤100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

1
2

484
2

Sample Input 2:

69 3

Sample Output 2:

1
2

1353
3

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个数，判断是否为回文数。如果不是，将其翻转+原数，得到一个新数，判断是否是回文数，如此循环。问N步之内是否可以得到回文数。

分析

因数据可能很大，采用string实现大整数加法。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;

bool ispal(string s1){
    string s2=s1;
    reverse(s2.begin(),s2.end());
    return s1==s2?true: false;
}

string flip(string s1){
    string s2=s1;
    reverse(s2.begin(),s2.end());
    int l=s1.length();
    int flag=0;
    for(int i=l-1;i>=0;i--){
        int t1=s1[i]-'0';
        int t2=s2[i]-'0';
        int t3=t1+t2+flag;
        if(t3>=10){
            flag=1;
            t3=t3%10;
        }
        else
            flag=0;
        s1[i]=t3+'0';
    }
    if(flag==1)
        s1="1"+s1;
    return s1;
}

int main() {
    string n;
    int k,i;
    cin>>n>>k;
    for(i=0;i<k;i++){
        if(ispal(n)){
            cout<<n<<endl;
            cout<<i;
            return 0;
        }
        else{
            n=flip(n);
        }
    }
    cout<<n<<endl;
    cout<<i;
    return 0;
}

五、其他

使用long int可过一部分测试点。

当输入196，100时，数据极大，必须使用string。