CCF认证 2019-3 小中大

水题

#include <iostream>
using namespace std;

int n;

int Max, Min, Middle;

double Middle1;

int main() {
	scanf("%d",&n);
	
	int arr[n];
	
	for(int i = 0; i < n; ++i) {	
		scanf("%d",&arr[i]);
	}
	
	Max = arr[0];
	Min = arr[n-1];
		
	if(Max < Min) {
		Max = arr[n-1];
		Min = arr[0];
	}
		
	if(n % 2 == 0) {
		if((arr[n/2-1] + arr[n/2]) % 2 == 0){
			Middle = (arr[n/2-1] + arr[n/2]) / 2;
			printf("%d %d %d",Max,Middle,Min);
		} else {
			Middle1 = (arr[n/2-1] + arr[n/2]) / 2.0;
			//.1f 用来四舍五入保留小数点后一位 
			printf("%d %.1f %d",Max,Middle1,Min);
		}
		
	}
	else {
		Middle = arr[n/2];
		printf("%d %d %d",Max,Middle,Min);
	}
	
	
	
	
	return 0;
}

标签：arr,小中大,Min,int,Max,Middle,Middle1,2019,CCF
来源： https://blog.csdn.net/shadiao_boy/article/details/121388934