Answer:
here
Explanation:
Python is an interpreted high-level general-purpose programming language. Its design philosophy emphasizes code readability with its use of significant indentation.
Answer:
#include <stdio.h>
#include <string.h>
int main(void) {
char simonPattern[50];
char userPattern[50];
int userScore;
int i;
userScore = 0;
scanf("%s", simonPattern);
scanf("%s", userPattern);
for(i = 0;simonPattern[i]!='\0';i++){
if(simonPattern[i]!=userPattern[i]){
userScore=i;
break;
}
}
printf("userScore: %d\n", userScore);
return 0;
}
Explanation:
- Use a for loop that runs until it does not reach the end of simonPattern.
- Check whether the current index of simonPattern and userPattern are not equal and then assign the value of i variable to the userScore variable and break out of the loop.
- Finally display the user score.
Yes it only processes data using 1s and 0s
A .) Sort
more ... To arrange or group in a special way (such as by size, type or alphabetically).
Answer:
a)
#include <iostream>
using namespace std;
int main() {
bool a,b,c;
cin>>a>>b;
if(a^b)//X-OR operator in C++.
c=true;
else
c=false;
cout<<c;
return 0;
}
b)
#include <iostream>
using namespace std;
int main() {
bool a,b,c,d;
cin>>a>>b>>c;
if((a^b)^c)//X-OR operator in C++.
d=true;
else
d=false;
cout<<d;
return 0;
}
Explanation:
The above written programs are in C++.There is an operator (^) called X-OR operator in C++.It returns true if the number of 1's are odd and returns false if the number of 1's are even.
In the if statement I have user X-OR operator(^) to find the result and storing the result in another boolean variable in both the questions.
