Answer:
Replace /* Your solution goes here */ with:
cin>>matchValue;
numMatches = 0;
for (i = 0; i < userValues.size(); ++i) {
if(matchValue == userValues.at(i))
{
numMatches++;
}
}
Explanation:
This line gets input for matchValue
<em>cin>>matchValue;
</em>
This line initializes numMatches to 0
<em>numMatches = 0;
</em>
The following iteration checks for the number of matches (numMatches) of the matchValue
<em>for (i = 0; i < userValues.size(); ++i) {
</em>
<em>if(matchValue == userValues.at(i))
</em>
<em>{
</em>
<em> numMatches++;
</em>
<em>}
</em>
<em>}
</em>
<em>See Attachment for full source code</em>
Answer:
The statement is as follows:
print("{0:,.1f}".format(number))
Explanation:
Required
Statement to print 1234567.456 as 1,234,567.5
To do this, we make use of the format keyword, and we set the print format in the process.
To round up number to 1 decimal place, we use the following format:
"{0:,.1f}"
To include comma in the thousand place, we simply include a comma sign before the number of decimal place of the output; i.e. before 1
"{0:,.1f}"
So, the print statement is:
print("{0:,.1f}".format(number))
Answer:
Please kindly check explainations for the code.
Explanation:
lw $t1, Num1
lw $t2, Num2
lw $t3, Num3
blt $t1, $t2, if
beq $t1, $t2, elseif
else:
add $t0, $t3, 5
sw $t0, Result
endif:
#.....other statements after if-elseif-else
if:
sw $t1, Result
b endif
elseif:
ble $t2, $t3, if2
or $t0, $t1, $t3
sw $t0, Result
b endif
if2:
and $t0, $t2, $t3
sw $t0, Result
b endif
Go to attachment for the onscreen code.
It is a class C address. Class C covers all addresses in the range 192.0.0.0 to 223.0.0.0. The /24 represents the number of network bits in the address so we can work out that the subnet mask of this address would be 255.255.255.0.
Answer:
def prime_generator(s, e):
for number in range(s, e+1):
if number > 1:
for i in range(2, number):
if (number % i) == 0:
break
else:
print(number)
prime_generator(6,17)
Explanation:
I believe you want to ask the prime numbers between s and e.
- Initialize a for loop that iterates from s to e
- Check if the number is greater than 1. If it is, go inside another for loop that iterates from 2 to that number. If the module of that number to any number in range (from 2 to that number) is equal to 0, this means the number is not a prime number. If the module is not equal to zero, then it is a prime number. Print the number
