Consider the C program given below :
#include <stdio.h> int main () {
int sum = 0, maxsum = 0, i, n = 6;
int a [] = {2, -2, -1, 3, 4, 2};
for (i = 0; i < n; i++) {
if (i == 0 || a [i] < 0 || a [i] < a [i - 1]) {
if (sum > maxsum) maxsum = sum;
sum = (a [i] > 0) ? a [i] : 0;
}
else sum += a [i];
}
if (sum > maxsum) maxsum = sum ;
printf ("%d\n", maxsum);
} |
What is the value printed out when this program is executed?
(A) 9
(B) 8
(C) 7
(D) 6
Answer: (C)
Explanation: If you look for loop carefully, you will notice that it assigns sum variable to some value in if condition and increments it in the else condition. On further thought, it would be clear that this loop stores sum of increasing subsequence of positive integers in sum variable and max of sum in maxsum. Hence, maxsum – maximum sum of increasing subsequence of positive integers will get printed out when this program is executed, which is 3 + 4 = 7.
This solution is contributed by Vineet Purswani
//output will be 3+4 =7 {for || if 1st argument is true 2nd argument will not be calculated, and if 1st argument is false, 2nd argument will be calculated}
Another Solution
When i=1
-> i==0 is false, but a[i] maxsum) is true, since sum=2 and maxsum=0.So maxsum=2. sum = (a [i] > 0) ? a [i] : 0; , sum=0 since a[i]<0.
When i=2
-> i==0 is false, a[i] maxsum) is false, since sum=0 and maxsum=2.Since sum = (a [i] > 0) ? a [i] : 0; , sum=0 since a[i]<0.
When i=3
-> i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 3.
When i=4
-> i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 7.
When i=5
-> i==0 is false , a[i]<0 is false and a [i] maxsum is true, since sum=7 and maxsum=2,so maxsum=7.Since sum =
(a [i] > 0) ? a [i] : 0, so sum=2 since a[5]>0.
This solution is contributed by nirmal Bharadwaj
Quiz of this Question