Huawei 2018 school recruit technical post written

输入一个整数（含负数），输出3个数据，如下：
1.输出该整数的位数;
2.将该整数各位拆分输出，中间以空格隔开（注意末位不能有空格）。如果是负数，则符号与第一个数一起输出；
3.输出该数的反转数，如为负数，符号位置不变，置于最前。

Example
Input:

-12345

Output:

 

5
-1 2 3 4 5
-54321

#include <the iostream> 
#include <the cmath> 
#include <algorithm> 
#include <SET> 
#include <cstdio> 
#include <String> 
#include <CString> 
/ * @ author: bk 
* Family: 
* Time: 
* / 
/ / I'm like a child playing at the beach, 
// time to time to pick up more than the usual smooth stones or more beautiful shells rejoice, 
// and demonstrated in front of me is not entirely proven sea truth 
using STD namespace; 

int main () 
{ 

    String s_num; 
    int I; 
    getline (CIN, s_num); 
    IF (s_num [0] == '-') negative // 
    { 
        COUT << s_num.size () -. 1 << endl ; 
        COUT << "-" << s_num [. 1] << ""; 
        for (I = 2; I <s_num.size ();i++){i++){
            cout<<s_num[i]<<(i==s_num.size()-1? "\n":" ");
        }
        cout<<"-";
        for(int i=s_num.size()-1;i>=1;i--){
            cout<<s_num[i];
        }


    }
    else{
       cout<<s_num.size()<<endl;
        for(i=0;i<s_num.size();i++){
            cout<<s_num[i]<<(i==s_num.size()-1? "\n":" ");
        }

        for(i=s_num.size()-1;i>=0;i--){
            cout<<s_num[i];
        }


    }



    //cout << "Hello world!" << endl;
    return 0;
}

　

topic:

输入4个IP值组成两个IP段：
第一、二行分别为第一个IP段的起始和结尾IP，第三、四行为第二个IP段的起始和结尾。
要求输出：
若两个IP段有交集则输出"Overlap IP"，没有则输出"No Overlap IP"。

Example
Input:

1.1.1.1
255.255.255.255
2.2.2.2
3.3.3.3

Output:

Overlap IP