太简单了，我花了整整3秒才想出来

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n, a[105], maxn = -1, minn = 999999,pos1,pos2;
    cin >> n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        if(a[i]>maxn)
        {
            maxn = a[i];
            pos1 = i;
        }
        if(a[i]<minn)
        {
            minn = a[i];
            pos2=i;
        }
    }
    int t = a[pos1];
    a[pos1] = a[pos2];
    a[pos2] = t;
    for(int i=1;i<=n;i++)
        cout << a[i]<<" ";
    return 0;
}

看不懂另一个

#include <iostream>
using namespace std;
int n,a[104];
int main()
{
    int max=-45,min=99999;
    cin >>n;
    for(int i=1;i<=n;i++)
    {
        cin >> a[i];
        if(a[i]>max)
        {
            max=a[i];
        }
        if(a[i]<min)
        {
            min=a[i];
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]==max)
        {
            cout << min<<" ";
        }
        else if(a[i]==min)
        {
            cout <<max<<" ";
        }
        else
        {
            cout <<a[i]<<" ";
        }
    }
}

这道题其实只需要用指针记录max和min的地址。然后循环输出，如果a[i]的地址与max或min的地址相同，就输出题目要求的对应的值。 AC Code

#include <bits/stdc++.h>
using namespace std;
int m,a[1000],maxx=INT_MIN,minn=INT_MAX;
int main(){
	cin>>m;
	int *x=NULL,*n=NULL;
	for(int i=1;i<=m;i++){
		cin>>a[i];
		if(a[i]>maxx){
			maxx=a[i];
			x=&a[i];
		}
		if(a[i]<minn){
			minn=a[i];
			n=&a[i];
		}
	}
	for(int i=1;i<=m;i++){
		if(&a[i]==x)cout<<minn<<" ";
		else if(&a[i]==n)cout<<maxx<<" ";
		else cout<<a[i]<<" ";
	}
	return 0;
}