Answer / abdur rab

for ( i = 0; i < 100; i++ )
for ( j = 100; j > 100 - i; j--)
sum++;

first iteration i = 0
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 0 = 100 )

so output is 0 ( sum is incremented 0 times )

second iteration i = 1
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 1 = 99 )

so output is 1 ( sum is incremented 1 times )

third iteration i = 2
j = 100
j loop is executed untill ( 100 - i ) (ie 100 - 2 = 98 )

so output is 1 ( sum is incremented 2 times )


0 + 1 + 2 + 3.......+ 99 = ( n (n+1) ) / 2

( 99 (99+1) ) / 2 = 4950

Answer / anil

0 times bcoz everytime it enters second loop condition is
not satisfied ,thus comes out of loop.

Answer / daniel

(99 * 100)/2 = 4950
The sum++ is performed 4950 times.

Answer / santosh

when i=0
j=100 and 100>100-1(false) come out of the loop and the sum
is executed 0 times

Answer / lnk

Its quite simple to analyse ...

LOOP i=0: i=0
then enters loop j=o: but it false that always J>100-i;

i.e.., i=0;j=100; 100- i-> 100
so always 100 is not greater than 100

than it comes out no sum++;
Loop i=1; ; j =100 only now 100 - i =99 so j>100 -i
(100>99)
then sum++ is executed ;


i= 1 j=100 j > 100 - i j=99 ;sum ++
i=2 j =99 j> 100 - i j =98 ; sum ++

i=50 j=51 j>100-50 true ( 51>50 ) ; sum++ j=50
i=51 j= 50 j>100-51 true(50>49 ) so no sum++


i = 99 j=2 j>100-i true (2>1) sum ++

so sum++ would be executed 99 times

what will be the output of this program? #include<stdio.h> #define cube(x) x*x*x void main() { int i,j=5; i=cube(j+3); printf("i=%d",i); }

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