A is correct! We did this in Law!
Have a merry Christmas!
Highway:
<span>cout << "The car can drive " << 20*26.8 << " miles on the highway." << endl; </span>
<span>Town: </span>
<span>cout << "The car can drive " << 20*21.5 << " miles in the town." << endl;</span>
If you save the input as num,
this will print the input 8 times.
num = input("Enter a number: ")
print(num * 8)
If you want to do actual math calculations,
then the input needs to be a number.
num = float(input("Enter a number: "))
print(num * 8)
This doesn't account for any errors in which the user doesn't input a number, but I don't think that's what you were looking for anyway :)
Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
