【c++】输入一个n×n的矩阵，求出两条对角线元素值之和，输出矩阵，求最大值、最小值及下标

#include<iostream>
using namespace std;

int main(){
    int n;
    int a[100][100],i,j;
    cout<<"请输入n*n矩阵的n值：";
    cin>>n;
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            cout<<"a["<<i<<"]["<<j<<"]=";
            cin>>a[i][j];
            
        }
        cout<<endl;
    }
    //输出矩阵
    cout<<"生成的矩阵为:\n";
    for(i=0;i<n;i++){
    for(j=0;j<n;j++){
        cout<<a[i][j]<<'\t';    
    }
    cout<<endl;
    }
    //求矩阵对角线和
    cout<<"矩阵对角线和为:";
    int sum=0;
    for(i=0;i<n;i++){
    for(j=0;j<n;j++){
        if(i==j){
            cout<<a[i][j];
            if (i<n-1) cout<<"+";
            else cout<<"=";
            sum+=a[i][j];
        }    
    }
    }
    cout<<sum<<endl;    
    //求矩阵最大值、最小值及下标
    int max,min,a1,a2,b1,b2;
    max=min=a[0][0];
    a1=a2=b1=b2=0;
    for(i=0;i<n;i++){
    for(j=0;j<n;j++){
        if (max<a[i][j]) {max=a[i][j];a1=i;b1=j;}
        if (min>a[i][j]) {min=a[i][j];a2=i;b2=j;}

    }
    }
    cout<<"矩阵最大值a["<<a1+1<<"]["<<b1+1<<"]="<<max<<endl;
    cout<<"矩阵最小值a["<<a2<<"]["<<b2<<"]="<<min<<endl;




    return 0;


}

输出结果为：