CZH_xixi @ 2024-04-11 15:34:17

#include<bits/stdc++.h>
using namespace std;
double n,a[1005],sum,ma,mi;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        sum+=a[i];
        ma=max(ma,a[i]);
        mi=min(mi,a[i]);
    }
    printf("%.2lf",(sum-ma-mi)/(n-2));
}

by 违规用户名1284771 @ 2024-04-11 16:20:33

你可以用cout<<mixed<<setprecision(2); 试试小数位数

by 违规用户名1284771 @ 2024-04-11 16:24:53

试试这个

https://www.luogu.com.cn/paste/ce4mwbev

by 违规用户名1284771 @ 2024-04-11 16:25:17

@2023_1202

by __Function__ @ 2024-04-21 17:28:04

你可以把ma与mi设成极端数据，如ma=8445579834，mi=-2387548（各种都可以）

by LiXiang190208 @ 2024-05-12 12:08:32

#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 anyiming @ 2024-05-22 23:05:18

@LiXiang190208 你的for循环的条件的位置写反了。应该是

for(int i=0; i<n; i++){
        cin>>a[i];
        sum+=a[i];
    }
for(int i=0; i<n; i++){
  s = max(s,a[i]);
  j = min(j,a[i]);
}

这样应该行了

by LiXiang190208 @ 2024-05-24 20:58:59

@ anyiming
Thank you!!

着急不小心出错o(╥﹏╥)o