chenqile @ 2023-12-22 19:21:24
代码如下
#include <bits/stdc++.h>
using namespace std;
int main(){
double a[1002], n, sum = 0;
cin>>n;
for (int i = 0; i < n; i++) cin>>a[i];
sort(a, a + n);
for (int i = 1; i < n - 1; i++) sum += a[i];
printf("%.1lf", sum/n);
return 0;
}
by xiaoshumiao @ 2023-12-22 19:24:59
@chenqile 仔细读题,他让你保留两位小数。
by 7The_shy @ 2023-12-22 19:27:21
#include <bits/stdc++.h>
using namespace std;
int main(){
double a[1002],sum = 0;
int n;
cin>>n;
for (int i = 0; i < n; i++) cin>>a[i];
sort(a, a + n);
for (int i = 1; i < n - 1; i++) sum += a[i];
printf("%.2lf", sum/(n-2));
return 0;
}
by chenqile @ 2023-12-22 19:28:45
怎么还是ce
by 7The_shy @ 2023-12-22 19:28:56
1.n不能为double
2.输出的时候n应该减去两人
3.保留两位应该是“%.2lf”
by xiaoshumiao @ 2023-12-22 19:29:34
@chenqile ?IDE 上没有 CE。
by xiaoshumiao @ 2023-12-22 19:30:23
@7The_shy 不过 n 硬要开 double 也没事。
by 7The_shy @ 2023-12-22 19:31:12
不过本地会报错
by chenqile @ 2023-12-22 19:34:35
THANKS EVERYBODY
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