matrix o matrix

[keith katthy]

i have done this programming, but when the output appears I must enter any number, until 6,but i dont want to make it repeat again?how can we avoid it?

then i want to make a random number for [i][j], but i dont know how to do?
Anyone?

#include <iostream>
#include <conio.h>
#include <cstdlib> 
#include <math.h>
using namespace std;
 
 
int  main()
{
    void srand ( unsigned int seed );
    int a[6][6] ;
    int i, j;
    int x[10];
    int A[10], B[10], C[10], D[10], E[10], F[10];
    int clrscr();
 
{
    for(i=1;i<7;i++)
    for(j=1;j<7;j++)
{
    for(i=1;i<7;i++)
{
    x[i]=i;
    A[i]=(2+pow(x[i],2));
    B[i]=(1+pow(x[i],3));
    C[i]=(2+pow(x[i],2)-1);
    D[i]=(3*pow(x[i],3));
    E[i]=(1+pow(x[i],2));
    F[i]=(3+pow(x[i],2));
}
{
    printf("\n");
 
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] A[%d]=%d\n", i,j, i, A[i]); 
}
    printf("\n");
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] B[%d]=%d\n", i,j, i, B[i]); 
 
} 
    printf("\n");
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] C[%d]=%d\n", i,j, i, C[i]); 
 
}
    printf("\n");
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] D[%d]=%d\n", i,j, i, D[i]); 
 
}
    printf("\n");
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] E[%d]=%d\n", i,j, i, E[i]); 
    }        
    printf("\n");
    for(i=1;i<7; i++)
 
{   
    printf("[%d][%d] F[%d]=%d\n", i,j, i, F[i]); 
}
 
    cin>>a[i] [j] ;
 
}
}
}
    for(i=1;i<7;i++)
{
    cout<<endl;
 
    printf("%d\t", A[i]);
    printf("%d\t", B[i]);   
    printf("%d\t", C[i]);
    printf("%d\t", D[i]);
    printf("%d\t", E[i]);
    printf("%d\t", F[i]);
}
    _getch();
 
 
    printf("\n");
 
    printf("Random number: %d\n", rand() % 100);
    srand ( 1 );
 
 
 
}

[sicarie]

I am extremely confused by your code. Why do you have all of the { and }'s? What do you mean when you say you 'don't want to make it repeat'? What don't you want to make repeat? The random number? Or are you assigning these yourself?

One option is to get the random number from rand() and then to a linear search of the array to see if it is in there yet. This will be very costly, time-wise, though.