Answer:
import random
numbers = []
even = 0
odd = 0
for i in range(100):
numbers.append(random.randint(1, 200))
for i in range(100):
if numbers[i] % 2 == 0:
even += 1
else:
odd += 1
print("Even:", even)
print("Odd:", odd)
Explanation:
Gg ez.
Answer:
A cell grows to its full size, The cell copies its DNA
have a great weekends, hopefully it was the right answer!
Answer:
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string str)
{
char a,b;
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
a=tolower(str[i]);//Converting both first characters to lowercase..
b=tolower(str[length-1-i]);
if (b != a )
return false;
}
return true;
}
int main() {
string t1;
cin>>t1;
if(isPalindrome(t1))
cout<<"The string is Palindrome"<<endl;
else
cout<<"The string is not Palindrome"<<endl;
return 0;
}
Output:-
Enter the string
madam
The string is Palindrome
Enter the string
abba
The string is Palindrome
Enter the string
22
The string is Palindrome
Enter the string
67876
The string is Palindrome
Enter the string
444244
The string is not Palindrome
Explanation:
To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.
<span>Not a valid IPv6 address
A valid IPv6 address consist of 8 groups of 4 hexadecimal numbers separated by colons ":". But that can make for a rather long address of 39 characters. So you're allowed to abbreviate an IPv6 address by getting rid of superfluous zeros. The superfluous zeros are leading zeros in each group of 4 digits, but you have to leave at least one digit in each group. The final elimination of 1 or more groups of all zeros is to use a double colon "::" to replace one or more groups of all zeros. But you can only do that once. Otherwise, it results in an ambiguous IP address. For the example of 2001:1d5::30a::1, there are two such omissions, meaning that the address can be any of
2001:1d5:0:30a:0:0:0:1
2001:1d5:0:0:30a:0:0:1
2001:1d5:0:0:0:30a:0:1
And since you can't determine which it is, it's not a valid IP address.</span>
