The Ninth Hunan Collegiate Programming Contest (2013) Problem G

Problem G

Good Teacher

I want to be a good teacher, so at least I need to remember all the student names. However, there are too many students, so I failed. It is a shame, so I don't want my students to know this. Whenever I need to call someone, I call his CLOSEST student instead. For example, there are 10 students:

A ? ? D ? ? ? H ? ?

Then, to call each student, I use this table:

Pos	Reference
1	A
2	right of A
3	left of D
4	D
5	right of D
6	middle of D and H
7	left of H
8	H
9	right of H
10	right of right of H

Input

There is only one test case. The first line contains n, the number of students (1<=n<=100). The next line contains n space-separated names. Each name is either ? or a string of no more than 3 English letters. There will be at least one name not equal to ?. The next line contains q, the number of queries (1<=q<=100). Then each of the next q lines contains the position p (1<=p<=n) of a student (counting from left).

Output

Print q lines, each for a student. Note that "middle of X and Y" is only used when X and Y are both closest of the student, and X is always to his left.

Sample Input

10
A ? ? D ? ? ? H ? ?
4
3
8
6
10

Output for the Sample Input

left of D
H
middle of D and H
right of right of H

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The Ninth Hunan Collegiate Programming Contest (2013) Problemsetter: Rujia Liu Special Thanks: Feng Chen, Md. Mahbubul Hasan

  打好基础，顺序查找，没有什么高端算法，就是希望在某些情况下少用各种break ，多用函数return ，这样好一点，少犯些错误。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
int N ;
string str[108] ;
void gao(int id){
    if(str[id]!="?"){
         cout<<str[id]<<endl ;
         return  ;
    }
    int L ,R ;
    L=R=1 ;
    int left=id-1 ;
    int right=id+1 ;
    str[0]=str[N+1]="?" ;
    while(left>=1&&str[left]=="?"){
        R++ ;
        left-- ;
    }
    while(right<=N&&str[right]=="?"){
        L++ ;
        right++ ;
    }
    if(L==R&&str[left]!="?"&&str[right]!="?"){
        printf("middle of %s and %s
",str[left].c_str(),str[right].c_str()) ;
        return ;
    }
    if((str[left]!="?"&&str[right]!="?"&&L>R)||str[right]=="?"){
        for(int i=1;i<=R;i++)
            printf("right of ") ;
        cout<<str[left]<<endl  ;
        return  ;
    }
    if((str[left]!="?"&&str[right]!="?"&&L<R)||str[left]=="?"){
        for(int i=1;i<=L;i++)
            printf("left of ") ;
        cout<<str[right]<<endl  ;
        return  ;
    }
}
int main(){
   int M ,id ;
   cin>>N ;
   for(int i=1;i<=N;i++)
       cin>>str[i] ;
   cin>>M ;
   while(M--){
       cin>>id ;
       gao(id) ;
   }
   return 0 ;
}