Question

用java编程一个N*N的矩阵，对角线上的元素为一，其他元素均为0.

Answer 1

public class C {
public static void main(String[] args) {
int n = 10;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i==j) {
System.out.print(1+" ");
}else if((i+j)==(n+1)){
System.out.print(1+" ");
}else{
System.out.print(0+" ");
}
}
System.out.println("");
}
}
}

Answer 2

int n = 100;
int i[][] = new int[n][n];
for(int j=0;j<n;j++){
    for(int k=0;k<n;k++){
        if((j==0||j==(n-1))&&(k==0||k==n-1)){
            i[j][k]=1;
        }else{
            i[j][k]=0;
        }
    }
}
System.out.println(i);

Answer 3

int N = 10;
int a[][] = new int[N][N];
for(int i=0;i<N;i++){
for(int ii=0;ii<N;ii++){
if(ii == i || ii == N-i-1) a[i][ii] = 1;
else a[i][ii] = 0;
}
}
//显示
for(int[] a1 : a){
for(int a2:a1){
System.out.print(a2+" ");
}
System.out.println();
}