你可以用cout<<mixed<<setprecision(2);
试试小数位数
by 违规用户名1284771 @ 2024-04-11 16:20:33
试试这个
https://www.luogu.com.cn/paste/ce4mwbev
by 违规用户名1284771 @ 2024-04-11 16:24:53
@[2023_1202](/user/1255341)
by 违规用户名1284771 @ 2024-04-11 16:25:17
你可以把ma与mi设成**极端数据**,如ma=8445579834,mi=-2387548(各种都可以)
by fangchenling @ 2024-04-21 17:28:04
```
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,a[1000],s=1,j=1000,sum=0;
cin>>n;
for(int i=0;i++;i<n){
cin>>a[i];
sum+=a[i];
}
for(int i=0;i++;i<n){
s = max(s,a[i]);
j = min(j,a[i]);
}
sum = sum-s-j;
cout<<fixed<<setprecision(2)<<sum*1.0/(n-2);
return 0;
}
```
## 大佬们,为什么只能输入n
# 新手哟!!
by LiXiang190208 @ 2024-05-12 12:08:32