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
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[] ;
void gao(int id){
if(str[id]!="?"){
cout<<str[id]<<endl ;
return ;
}
int L ,R ;
L=R= ;
int left=id- ;
int right=id+ ;
str[]=str[N+]="?" ;
while(left>=&&str[left]=="?"){
R++ ;
left-- ;
}
while(right<=N&&str[right]=="?"){
L++ ;
right++ ;
}
if(L==R&&str[left]!="?"&&str[right]!="?"){
printf("middle of %s and %s\n",str[left].c_str(),str[right].c_str()) ;
return ;
}
if((str[left]!="?"&&str[right]!="?"&&L>R)||str[right]=="?"){
for(int i=;i<=R;i++)
printf("right of ") ;
cout<<str[left]<<endl ;
return ;
}
if((str[left]!="?"&&str[right]!="?"&&L<R)||str[left]=="?"){
for(int i=;i<=L;i++)
printf("left of ") ;
cout<<str[right]<<endl ;
return ;
}
}
int main(){
int M ,id ;
cin>>N ;
for(int i=;i<=N;i++)
cin>>str[i] ;
cin>>M ;
while(M--){
cin>>id ;
gao(id) ;
}
return ;
}