PAT(Basic Level) Practice : 1079 延迟的回文数 (20分)

1079 延迟的回文数 (20分)

测试点234

测试点234是输入就是回文数的情况，特殊判断即可。

代码

#include <iostream>
#include <vector>
#include <string>
#include <cstdio>
//scanf printf防止超时
#include <algorithm>
//vector的sort
#include <sstream>
//转换
using namespace std;

#include<iomanip>
//精度

#include<cmath>
//round四舍五入取整

bool judge(string str)
{
    int mid=str.length()/2;
    bool flag=true;
    for(int i=0;i<mid;i++)
    {
        if(str[i]!=str[str.length()-1-i])
            flag=false;
    }
    return flag;
}



int main()
{
    string str;
    cin>>str;
    if(judge(str))
    {
        cout<<str<<" is a palindromic number."<<endl;
    }else{
    for(int i=0;i<10;i++)
    {
        cout<<str<<" + ";
        for(int j=str.length()-1;j>=0;j--)
        {
            cout<<str[j];
        }
        cout<<" = ";

        int flag=0;
        string res="";
        for(int j=0;j<str.size();j++)
        {
            int sum=str[j]-'0'+str[str.size()-1-j]-'0'+flag;
            if(sum>=10)
            {
                sum-=10;
                flag=1;
            }else
            {
                flag=0;
            }
            char temp=sum+'0';
            res=temp+res;
        }
        if(flag==1)
            res="1"+res;
        cout<<res<<endl;
        if(judge(res))
        {
            cout<<res<<" is a palindromic number."<<endl;
            str=res;
            break;
        }
        str=res;
    }}

    if(!judge(str))
    {
        cout<<"Not found in 10 iterations."<<endl;
    }
    return 0;
}