Given the source code below: #include using namespace std; double f(double a[], int 1, int r)...

 

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Given the source code below: #include using namespace std; double f(double a[], int 1, int r) { } if (1=1) return a[1]: int m = (1+r)/2; double u = f(a.l.m); double v = f(a.m+1,r); return (u+v)/2; int main() { double a[] = {10,20,30,40,50,60,70,80}; int n = sizeof(a)/sizeof(double); cout < Given the source code below: #include using namespace std; double f(double a[], int 1, int r) { } if (1=1) return a[1]: int m = (1+r)/2; double u = f(a.l.m); double v = f(a.m+1,r); return (u+v)/2; int main() { double a[] = {10,20,30,40,50,60,70,80}; int n = sizeof(a)/sizeof(double); cout <

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Answer a The output of the algorithm for the given input in the main function will be the average of ... View the full answer