Look at your code first:
# contains "stdio.h"
void main(){
int a[6][6]，I，j；
for(I = 1； I<6; i++){
for(j = 1； j & lt6; j++){
a[I][j]=(I/j)*(j/I)；
}
}
for(I = 1； I<6; i++)
{
for(j = 1； j & lt6; j++){
printf("%2d "，a[I][j])；
}
printf(" \ n ")；
}
System ("suspended");
}
This is the result of the operation, you really choose C.
Next analysis, the core code is only this piece:
for(I = 1； I<6; i++){
for(j = 1； j & lt6; j++){
a[I][j]=(I/j)*(j/I)；
}
}
/*
When I =1:(1) j =1a [1] =1.
(2)j = 2a[ 1][2]= 0； Note: I and J are both i< types, so when I
There is no need to analyze i<: as long as I
Look at me>J (j/i) is also zero. So a[i][j] is also equal to zero.
When I = J, A[i][j] is equal to 1.
In fact, the logic is quite simple, mainly because I and J are both int types, and this A [I] [J] = (I/J) * (J/I); The equation has both i/j and J/I.
So whatever I do,
*/
Output also starts from a[ 1][ 1], so a [1] [1] =1; a[2][2]= 1； a[3][3]= 1； a[4][4]= 1； a[5][5]= 1；
It is the result of the above picture.