【PAT】1024. Palindromic Number (25)

1、題目地址：http://www.patest.cn/contests/pat-a-practise/1024

2、分析

在做這答題的過程中很可能出現下面的思路：

#include "stdio.h"


long long reverse(long long n)
{
    long long m = 0;
    long long tmp = 10;
    do{

        m = m*10 +  n % tmp;
        n = n/tmp;
    }while(n);

    return m;

}

int main()
{
    long long n;
    int k;
    scanf("%lld%d",&n,&k);
    long long m;
    int i = 0;
    if(n == reverse(n)){
        ;
    }else {
        for(i = 1;i <= k;++i){
            m = reverse(n);
            n += m;
            if(n == reverse(n))
                break;
        }
    }
    if(i > k)
        i = k;
    printf("%lld\n%d\n",n,i);
}

但提交時，會出現不能全部通過，這是因爲：原始的N最多是10^10,但在後來的翻轉過程中，N會變得很大，但小於等於(10^10)*(2^100)

3正確代碼：

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

string add(const string &s1, const string &s2)
{
    int sum,carry = 0;
    char digit;
    string ans;
    int i = s1.size() - 1 , j = s2.size() - 1;
    for(;i >= 0 && j >= 0;--i,--j){
        sum = s1[i] - '0' + s2[j] - '0' + carry;
        digit = sum % 10 + '0';
        ans.push_back(digit);
        carry = sum / 10;

    }
    int k = i < 0 ? j : i;
    const string &str = i < 0 ? s2 : s1;
    while(k >= 0){
        sum = str[k] + carry;
        digit = sum % 10 + '0';
        carry = sum / 10;
        ans.push_back(digit);
    }
    if(carry > 0){
        digit = carry + '0';
        ans.push_back(digit);
    }
    reverse(ans.begin(),ans.end());
    return ans;
}




int main()
{
    string str;
    int k;
    cin >> str >> k;


    string res = str; 
    reverse(res.begin(),res.end());
    int i;
    for(i = 0;i < k;++i){
        if(str == res)
            break;
        str = add(str,res);
        res = str;
        reverse(res.begin(),res.end());
    }
    cout << str << endl << i << endl;

}