coderzhx @ 2024-06-06 21:25:52
#include <iostream>
using namespace std;
long p = -66666;
long q = 88888;
double a[5];
double n;
double sum;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
sum += a[i];
if (a[i] > p) {
p = a[i];
}
if (a[i] < q) {
q = a[i];
}
}
printf("%.2lf", (sum - (p + q)) / (n - 2) * 1.0);
return 0;
}
by _czx6666_ @ 2024-06-06 21:27:07
@Geyijinnoip 数据范围不看吗?
double a[5000];
by _czx6666_ @ 2024-06-06 21:27:23
求关
by coderzhx @ 2024-06-06 21:32:01
@czx6666 哦,谢谢,A了啊