``````int main ()
{
int numb[10];//1,2,4,8,16,32,64,128,256,512
int numb2[10];
int counter, counter2;

/*
numb[0]=1,numb[1]=2,numb[2]=4,numb[3]=8,numb[4]=16,numb[5]=32;
numb[6]=64,numb[7]=128,numb[8]=256,numb[9]=512;

numb2[0]=0,numb2[1]=1,numb2[2]=5,numb2[3]=3,numb2[4]=10,numb2[5]=5;
numb2[6]=15,numb2[7]=7,numb2[8]=20,numb2[9]=9;
*/
for (int z=0; z<10; z++)
cout<<numb[z]<<" ";
cout<<endl<<endl;

for (int x=0; x<10; x++)
cout<<numb2[x]<<" ";
cout<<endl<<endl;

return 0;
}``````
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Discussion Span
Last Post by timdog345

Initializing numb is easy. Just start at 1 and double the last value until you get to the size of the array:

``````int k = 1;

for (int i = 0; i < 10; ++i) {
numb[i] = k;
k *= 2;
}``````

Initializing numb2 is a lot harder. Edward can't find a good pattern in the current list of numbers. Alternating between adding by 5 and dividing by 2 is close, but far enough off that I'm not sure if that's what you're looking for.

Thank you

``````for( int i=0 ; i<10 ; ++i )
numb2[i] = i%2 == 0 ? (i/2)*5 : i ;``````

Thank You but this only helps with half

``````for( int i=0 ; i<10 ; ++i )
numb2[i] = i%2 == 0 ? (i/2)*5 : i ;``````

the numbers should be 0,1,5,3,10,5,15,7,20,9
even positions have values 0,5,10...
odd positions have values 1,3,5...