C :

#include<stdio.h>
int main(){
	int n,i,j,t,s=0;
	scanf("%d",&n);
	int a[n];
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
		s = s + a[i];
	}
	for(i=0;i<n-1;i++){
		for(j=0;j<n-1-i;j++){
			if(a[j]>a[j+1]){
				t = a[j];
				a[j] = a[j+1];
				a[j+1] = t;
			}
		}
	}
	int min = a[0],k=0;
	for(i=0;i<n;i++){
		if(a[i]==min){
			k++;
		}
	}
	printf("%.1lf",(s-k*min)*1.0/(n-k));
	return 0;
}

C++ :

#include <iostream>
#include <iomanip>
using namespace std;

int main(){
	int a[1000];
	int n,i,j,min,s = 0;
	cin>>n;
	
	for(i = 0;i < n;i++){
		cin>>a[i];
		s = s + a[i];
	}
	
	min = a[0];
	for(i = 0;i < n;i++){
		if(a[i] < min){
			min = a[i];
		}
	}
	
    int x = 0; 
    for(i = 0;i < n;i++){
		if(a[i] == min){
			x++; 
		}
	}
	
	double v = (s - x * min) * 1.0 / (n - x);
	cout<<fixed<<setprecision(1)<<v<<endl;
}

Python :

n = int(input())
m = input().split()
l = []
mi = 0
su = 0
c = 0
#列表类型转换
for v in m:
    l.append(int(v))
#求列表最小值
mi = min(l)
#遍历l列表,把列表中等于最小值的元素去掉,得出余下的值的和
for i in range(0, n):
    if l[i] != mi:
        su += l[i]
        c += 1
print('%.1f' % (su / c))