#include <bits/stdc++.h>
#define N 105 
using namespace std; 
 
int a[N], n; 
 
int main() { 
    cin>>n;
    int num=0;
    int result[n];
    for (int i =0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<n;i++)
    {
        for (int j=0;j<n;j++){
            for(int k=j+1;k<n;k++){
                if( (i!=j)&(j!=k)){
                    if(a[i]==a[j]+a[k]){
                        num+=1;
                       

                    }
                        

                }
            }
        }
    }
    
  
      cout<<num;
  
  
  
  
    return 0; 
}

#include <bits/stdc++.h>
#define N 105 
using namespace std; 

int a[N], n; 

bool check(int i) { 
        for (int j = 0; j < n; ++j) 
            for (int k = j + 1; k < n; ++k) 
                if (a[j] + a[k] == a[i]) return true; 
    return false; 

} 

 

int main() { 
    scanf("%d", &n); 
    for (int i = 0; i < n; ++i)  
        scanf("%d", &a[i]); 

    int sum = 0; 
    for (int i = 0; i < n; ++i) 
        if (check(i)) ++sum; // 也可以是 sum += check(i)

    printf("%d\n", sum); 
        return 0; 

}

 2排列枚举问题

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std ;

int main(){
    int a,b,c;
    cin>>a>>b>>c;
    int arr[10]={1,2,3,4,5,6,7,8,9,10};
    bool check=false;
    do{
        int num1=arr[0]*100+arr[1]*10+arr[2];
        int num2=arr[3]*100+arr[4]*10+arr[5];
        int num3=arr[6]*100+arr[7]*10+arr[8];
        if (a*num2==b*num1&&c*num2==b*num3){
            cout<<num1<<" "<<num2<<" "<<num3<<endl;
            check=true;
        }


    }while(next_permutation(arr,arr+9));
    if(!check){
        cout<<"No!!!"<<endl;
        
    }


  
  
  
  
  
  
    return 0 ;
}

 冒泡排序和选择排序有什么区别？？

插入排序怎么理解

 不是很理解，还是不行

#include <bits/stdc++.h>
using namespace std; 

int main() {
    // 请补全代码，实现题目功能
int n;
cin>>n;
int a[n];
for (int i=1;i<=n;i++){
    cin>>a[i];
}
for (int i=2;i<=n;i++){
    int j;int x=a[i];
    for(j=i;j>1&a[j-1]>a[j];j--){
        a[j]=a[j-1];

    }
//     a[j]=x;
}
sort(a+1,a+n+1);
for (int i=1;i<=n;i++){
    cout<<a[i]<<" ";
}
    return 0;
}