...inside that for loop, it will be outputted

**n**amount of times, because of the condition,**j**<**n**. consider this for ( int i =0 ; i <=5 ; i++ ) puts("hello...4 Answers · Computers & Internet · 13/11/2011

for (int i = 0; i < 2; i++) { for (int

**j**= 0;**j**< 2;**j**++) { res[i][**j**] = m[i][**j**] +**n**[i][**j**]; }} =) MJ3 Answers · Computers & Internet · 10/09/2007

... * if

**j**=0, and the array is strictly decreasing? * if**j**=**n**-1, and the array is strictly increasing? ....**but**I think the overall approach should work fine. @M1 Answers · Computers & Internet · 05/08/2013

...for(i=(

**n***(**n**+1)/2)-**n**,**j**=**n**; i =0; i -= (**j**-1),**j**--) { for(k=0; k<**j**; k++) { printf( %c , ( A +i+k)); } printf(**n**); } this should work,**but**check the indexes, i really get confused with those! let check with...5 Answers · Computers & Internet · 05/03/2014

...more "Hello" The outer loop is done again with

**j**= 3,**but**then m <**n**is in values 9 < 9 so the inner loop is not...1 Answers · Computers & Internet · 09/10/2010

First you get drafted. Then you get money

2 Answers · Computers & Internet · 28/05/2013

..., 1..

**n**, so: O(**n**^2) complexity O((**n**- a)^2)**but**effectively O(**n**^2) ==> O(**n**^2) d) for (i = 0; i <**n**; i ++) for (**j**= 0;**j**<**n**;**j**++) S; for (**j**= 0;**j**<...3 Answers · Computers & Internet · 23/04/2013

the inner loop (

**j**) does something**n**^2 times.**but**that loop is run by the outer loop (i)**n**/2 times. so it'd be something like (**n**^2)*(**n**/2) which is (**n**^3/2) the most significant part of that is**n**^31 Answers · Computers & Internet · 28/09/2010

..., I'm not sure if you want to do this for clarity reasons,

**but**your if statements are unnecessarily long. (((**n**-(**j**+k))+k) >**j**) && ((**n**-(**j**+k))+**j**)>k**n**-(**j**+k)+k is just**n**-**j**, and the ...2 Answers · Computers & Internet · 02/08/2013

int i,

**j**; int**n**= last - 'a'; char s[]="abcdefghijklmnopqrstuvwxyz"; for(i=0;i<=**n**;i++){ for(**j**=0;**j**<=**n**;**j**++){ cout << s[i]<< s...2 Answers · Computers & Internet · 10/11/2012