Good Teacher（模拟）-白红宇

Good Teacher（模拟）

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发布时间：2019-06-20

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Good Teacher

Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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

A ? ? D ? ? ? H ? ?

Sample Output

left of D

middle of D and H

right of right of H

题解：老师点名，由于一些同学不认识名字，所以要找离他最近的同学代替，题中的I call his CLOSEST student instead；

还有一点就是两个同学中间，但是这两个同学间不能有人。。。

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          using namespace std;string name[110];map
           
             ans;int p[110];bool js(int i){ if(ans.count(i) == 0) return true; else return false;}int main(){ int n, q; while(~scanf("%d", &n)){ int tp = 0; ans.clear(); for(int i = 1; i <= n; i++){ cin >> name[i]; if(name[i] != "?"){ ans[i] = name[i]; p[tp++] = i; } } for(int i = 0; i < tp - 1; i++){ if((p[i + 1] + p[i]) % 2 == 0 && js((p[i + 1] + p[i])/2)) ans[(p[i + 1] + p[i])/2] = "middle of " + name[p[i]] + " and " + name[p[i + 1]]; } for(int i = 1; i <= n; i++){ if(ans.count(i) == 0){ int x = lower_bound(p, p + tp, i) - p; int cnt, cnt1; if(x == 0){ cnt = p[0] - i; while(cnt--){ ans[i] = "left of " + ans[i]; } ans[i] += name[p[0]]; } else{ cnt = i - p[x - 1]; if(x < tp)cnt1 = p[x] - i; if(x == tp || cnt < cnt1){ while(cnt--){ ans[i] = "right of " + ans[i]; } ans[i] += name[p[x - 1]]; } else{ while(cnt1--){ ans[i] = "left of " + ans[i]; } ans[i] += name[p[x]]; } } } } int q, x; scanf("%d", &q); while(q--){ scanf("%d", &x); cout << ans[x] << endl; } } return 0;}