<u>Answer:</u>
<em>int fNumber,scndNumber = -1, </em>
<em>dup = 0;
</em>
<em>do {
</em>
<em>cin >> fNumber;
</em>
<em>if ( scndNumber == -1) {
</em>
<em>scndNumber = fNumber;
</em>
<em>}
</em>
<em>else {
</em>
<em>if ( scndNumber == fNumber )
</em>
<em>duplicates++;
</em>
<em>else
</em>
<em>scndNumber = fNumber;
</em>
<em>}
</em>
<em>} while(fNumber > 0 ); </em>
<em>cout << dup;
</em>
<u>Explanation:</u>
Here three variables are declared to hold the first number which is used obtain all the inputs given by the user, second number to hold the value of <em>last encountered number and “dup” variable to count the number of duplicate values.</em>
<em>“Do-while”</em> loop help us to get the input check whether it is same as previous input if yes then it <em>adds to the duplicate</em> value otherwise the new previous value if stored.
The best possible fit would be a person with a creative mind and flexible mind.
The technical stuff can be taught and learnt. But to evolve and keep up with the pace of how technology improves day to day, you'd need some with flexibility to cope up with this and creativeness to make innovations.
Answer:
The answer is that it is a speaker note.
Explanation:
It leaves a note for people that use presentation files. I use it all the time on my google slides.
Solution:
initial = float(eval(input('Enter the monthly saving amount: ')))
x = (1 + 0.00417)
month_one = initial * x
month_two = (initial + month_one) * x
month_three = (initial + month_two) * x
month_four = (initial + month_three) * x
month_five = (initial + month_four) * x
month_six = (initial + month_five) * x
print('The sixth month value is: '+str(month_six))
Don't forget the saving amount, and initialize the balance with that amount. Inside the loop, work out and add the interest and then add the saving amount for the next month.
balance = 801
for month in range(6):
balance = balance * (1.00417)
print(balance)
Answer:
Running RECURSIVE-MATRIX-CHAIN is asymptotically more efficient than enumerating all the ways of parenthesizing the product and computing the number of multiplications of each.
the running time complexity of enumerating all the ways of parenthesizing the product is n*P(n) while in case of RECURSIVE-MATRIX-CHAIN, all the internal nodes are run on all the internal nodes of the tree and it will also create overhead.
Explanation:
