试题编号: | 201903-1 |
试题名称: | 小中大 |
时间限制: | 1.0s |
内存限制: | 512.0MB |
问题描述:
解题思路:
解题代码:
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int a[100001],n,amax,amin,amid;
int main(){
scanf("%d",&n);
scanf("%d",&a[0]);
amax = amin = a[0];
for(int i = 1;i < n; i++){
scanf("%d",&a[i]);
amax = amax > a[i] ? amax : a[i];
amin = amin < a[i] ? amin : a[i];
}
if(n % 2 == 1) {
printf("%d %d %d", amax, a[n/2], amin);
} else {
if((a[n/2-1] + a[n/2]) % 2 == 1)
printf("%d %.1lf %d", amax, (double)((a[n / 2 - 1] + a[n / 2]) / 2.0), amin);
else
printf("%d %d %d", amax, (a[n / 2 - 1] + a[n / 2]) / 2, amin);
}
return 0;
}