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Many college students and recent grads destroy their resumes by accompanying them with halfhearted or downright terrible cover letters. While some employers don't bother reading cover letters, most do. And they will quickly eliminate you if you make these cover letter mistakes
Answer:
#include <iostream>
#include <cstring>
using namespace std;
void replacePeriod(char* phrase) {
int i = 0;
while(*(phrase + i) != '\0')
{
if(*(phrase + i) == '.')
*(phrase + i) = '!';
i++;
}
}
int main() {
const int STRING_SIZE = 50;
char sentence[STRING_SIZE];
strcpy(sentence, "Hello. I'm Miley. Nice to meet you.");
replacePeriod(sentence);
cout << "Updated sentence: " << endl;
cout << sentence << endl;
return 0;
}
Explanation:
- Create a function called replacePeriod that takes a pointer of type char as a parameter.
- Loop through the end of phrase, check if phrase has a period and then replace it with a sign of exclamation.
- Inside the main function, define the sentence and pass it as an argument to the replacePeriod function.
- Finally display the updated sentence.
Answer:
<u>Solution a</u>
- n = int(input("Enter an integer: "))
-
- sum = 0
-
- for x in range(1, n+1):
- sum += x
-
- print(sum)
<u>Solution b</u>
- n = int(input("Enter an integer: "))
-
- for a in range(1, n + 1):
- sum = 0
- for x in range(1, a+1):
- sum += x
- print(sum)
Explanation:
Solution code is written in Python 3.
<u>Solution a</u>
First get the user input for an integer (Line 1).
Create a variable sum and initialize it with zero value (Line 3).
Create a for loop to traverse through the number from 1 to integer n (inclusive) and sum up the current number x (Line 5-6).
Print the sum to terminal (Line 8).
<u>Solution b</u>
Create an outer for loop that traverse through the number from 1 to integer n (inclusive) (Line 3).
Create an inner loop that traverse through the number from 1 to current a value from the outer loop and sum up the current x value (Line 5-6).
Print the sum to terminal (Line 7) and proceed to next iteration and reset the sum to zero (Line 4).
Answer:
for (char i='a'; i<='e'; i++){
for (char j='a'; j<='e'; j++){
cout << i<< j<< "\n";
}
}
Explanation:
The loop runs all characters from the inner while the outer holds one character, by doing so, a will be matched with a,b,c,... Like wise b,c,d,... and on it goes.
