Hey all, I'm having a little trouble with a homework problem. Here it is:
Write a problem that creates the following pattern:
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
Now I know that this will involve nested for loops, but every attempt I've made gives me results like:
999999999
888888888
777777777
666666666
555555555
444444444
333333333
222222222
111111111
Any ideas?
make the second loop depend on how far the first loop's gone and make it count from 1-9 (and counting down) instead of just filling the array with which value the first loop was on.
They say that if you play a Windows Install CD backwords, you hear satanic messages. That's nothing; play it forward and it installs Windows.
this is what I have based on that...it's wrong , I know. Help.Code:#include <stdio.h> #include <stdlib.h> int main(void) { /*Local Definitions*/ int row; int col; int limit; /*Statements*/ limit = 9; for (row = 1; row < limit; row++) for (col = 1; col < limit; col--) printf("%d", row); printf("\n"); system("PAUSE"); return 0; }
melee
to clerify my first post I didn't mean that the code should be counting down using -- I ment that is should shorten how high the values would be able to go, sorry if if it caused you any trouble.
something like
Code:for (i=0:i<10;i++) { for (j=0;j<10-i;j++) { /*stuff*/ } /*stuff*/ }
Last edited by Shogun; 10-18-2004 at 03:01 AM.
They say that if you play a Windows Install CD backwords, you hear satanic messages. That's nothing; play it forward and it installs Windows.
Hi, try it likle this, it should produce the output as you're asked for.
Code:#include <stdio.h> int main(void) { int y=0, z=0; for(y=9; y>0; y--) { for(z=1; z<y+1; z++) { fprintf(stdout,"%d",z); /* print the numbers */ } /* end of nested for loop */ printf(stdout,"\n"); /* print the new line */ } /* end of outer loop */ return 1; }
Any occasion I can use this code makes for a good day!
Code:void O_Oy( char *o_o, int o_O ) { char V_V[BUFSIZ]="%s\n"; char T_T[BUFSIZ]="%%[^%s]"; char v_v[BUFSIZ]={0}; char x_x[BUFSIZ]={0}; while( o_O -- ) { sprintf( v_v, T_T, o_o + o_O + 1 ); sscanf( o_o, v_v, x_x ); printf( V_V, x_x ); } } int main( void ) { char o_o[] = "1234567890"; O_Oy( o_o, 0x9 ); return 0; }
Quzah.
Hope is the first step on the road to disappointment.
Quzah, I love that code...unfortunately, it is completely over my head at the moment.However, I'm saving it...and one day...when I'm a grown up programmer, I'll understand it...hopefully. Thanks!!
micklarko, shogun: Thank you ever so much. Shogun's code worked excellently!
melee
