这道题所有巨佬都忽视了一个问题，一个循环解决不香吗？？？直接上代码好吧

cin >> x;
if (maxx < x) maxx = x;
if (minn > x) minn = x;

这就是我的核心代码，具体代码参考奎若大佬方法一代码！

rth rwhhhhhhhhhhhhhhhhhhh

#include <stdio.h>//调用scanf() printf()
#include <algorithm>//调用sort()
using namespace std;
int a[3];//三个礼物，只需要a[3]即可，数组从a[0]开始
int main(){
	for(int i = 0 ; i < 3 ; i ++)
		scanf("%i", &a[i]);//输入
	sort(a,a+3);//排序，找最大最小
	printf("%i\n%.1f",a[0] + a[2],(a[0]+a[2])/2.0);
    //"%i"为格式化整型 "\n"为换行符 "%.if"为输出一位小数
    //除以2.0是强制转换int为实数类型
	return 0 ;
}

三个数中找出两个数（最大数和最小数）

比如这三个数是x,y,z,会有xy,xz,yz三种结果，这两个数分别是最大数和最小数,至于谁大谁小，我们每种组合里面又会分为两种情况，比如xy,可以是x最大，y最小;也可以是y最大,x最小

  if(x>y && y>z || z>y && y>x)//xz
  {
  	cout<<x+z<<endl<<fixed<<setprecision(1)<<(x+z)/2;
  }
  if(x>z && z>y || y>z && z>x)//xy
  {
  	cout<<x+y<<endl<<fixed<<setprecision(1)<<(x+y)/2;
  }
   if(y>x && x>z || z>x && x>y )//zy
  {
  	cout<<y+z<<endl<<fixed<<setprecision(1)<<(z+y)/2;
  }

cc

tr                                                                          r                                                               rt             rt

cc

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