nbut 1115, 乱搞？

B - Cirno's Trick

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit Status Practice NBUT 1115

Description

Cirno(チ ルノ) is an ice elf. Her hobby is tricking on others. But the biggest problem is that - Cirno is too foolish to do tricking. She knows this, so she makes statistics to calculate the probability that she can do tricking successfully.

Cirno writes down some probabilities that she may succeed. And she delete the lowest one, because it's too foolish and she can't believe in it. Then she delete the highest one, because she thinks it's impossible.
After that, she wants to know the average of the left probabilities.

Input

This problem has several cases.
The first line of each case is an integer N (2 < N <= 1000), indicates the number of probabilities.
Then follows one line which containing N integers, indicate the probabilities. Each probability is a decimal between 0 and 100.

Output

For each case, you should output the average of the left probabilities. Keep two decimal places.

Sample Input

3
0.0 50.0 100.0

Sample Output

50.00

Hint

Use 'double' to replace 'float'.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

int n;
double p[1111];

int main(){
//  fstream fin("C:\\Users\\Administrator\\Desktop\\in.txt",ios::in);

    while(scanf(" %d",&n)==1){
        int i,j;
        double s=0,t1=111,t2=0;
        for(i=0; i<n; i++){
            scanf(" %lf",&p[i]);
            if(t1 > p[i]) t1=p[i];
            if(t2 < p[i]) t2=p[i];
            s+=p[i];
        }
        s =s- t1-t2;
        printf("%.2f\n",s/(n-2));
    }

//  fin.close();
    return 0;
}