the i loop counting from 10 down to 1
the nested j loop would output i for i times
10 would be a special case, might require an if 10 than output 0
or an output last character each time to strip the leading 1.
Brilliant moves has brilliantly solved it in Java.
Loop condition will remain same in all languages, only syntax will differ.
Nested loops. Google should help
The questioner has already solved this one; for a look at an easy solution in
C, see http://pastebin.com/1UUqqi28 under the name numbs.c
>
> John (gnujohn@gmail.com)
for(int i=10; i>=1; i--)
{
for(int x=1; x<=i; x++)
print i%10
}
}
for (int i=10; i>=1; i--) {
for (int j=1; j<=i; j++) {
System.out.print(i%10);
} // end for j
System.out.println(); // new line
} // end for i
here is the logic:
int j,k;
for(j=10;j>=0;j--){
for(k=j;k>0 ;k--)
cout << j%10;
cout << endl;
}
Use for repetition statements and increment/decrement operators to produce the output.
0000000000
999999999
88888888
7777777
666666
55555
4444
333
22
1
It will decrement, but do not know how to go ahead
