Answer:
7
Explanation:
This one obtains results that the others in particular do not have, whether it is even one more word or another initial sentence
Answer:
// CPP program to Convert characters
// of a string to opposite case
#include<iostream>
using namespace std;
// Function to convert characters
// of a string to opposite case
void convertOpposite(string &str)
{
int ln = str.length();
// Conversion according to ASCII values
for (int i=0; i<ln; i++)
{
if (str[i]>='a' && str[i]<='z')
//Convert lowercase to uppercase
str[i] = str[i] - 32;
else if(str[i]>='A' && str[i]<='Z')
//Convert uppercase to lowercase
str[i] = str[i] + 32;
}
}
// Driver function
int main()
{
string str = "GeEkSfOrGeEkS";
// Calling the Function
convertOpposite(str);
cout << str;
return 0;
}
Explanation:
Answer:time-sensitive email
Answer:
Answer is B, see explanations.
Explanation:
Answer is (B) NP−complete∩P=ϕ
Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .
Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.
