Write an application named Perfect that displays every perfect number from 1 through 10,000. A number is perfect if it equals th
e sum of all the smaller positive integers that divide evenly into it. For example, 6 is perfect because 1, 2, and 3 divide evenly into it and their sum is 6.
1 answer:
Answer:
Required code is given and the output is also attached.
Explanation:
public class Perfect {
public static void main(String[] args) {
int i, num, sum;
System.out.println("The perfect numbers between 1 and 1000 are");
num = 1;
while (num <= 1000) {
sum = 0;
for (i = 1; i < num; i++)
if (num % i == 0)
sum += i;
if (sum == num)
System.out.println(num + " ");
num++;
}
}
}
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Answer:
Here is the C++ program:
#include<iostream> //to use input output functions
#include <algorithm> //to use tolower() function
using namespace std; //to identify objects like cin cout
int main() { //start of main function
string text; // to hold text input
cout<<"Enter a line of text: "; // prompts user to enter a string
getline(cin,text); // reads the input string (text) from user
transform(text.begin(), text.end(), text.begin(), ::tolower); //converts the text into lower case
int letter[26] = {0}; //to hold the letters
int i; // used as a loop variable
int words=0; // to hold the count of words
for(i = 0; i< text.size();i++){ // iterates through the text
if(isalpha(text[i])&&(text[i+1]=='.'||text[i+1]==','||text[i+1]==' ')) //checks if the character at i-th index of text is a letter, and checks if i+1 index position of text is a period, comma or a space
words++; // adds 1 to the count of words
if(isalpha(text[i])) // if the character at the i-th index of text is a letter
letter[text[i]-'a']++; } // counts the occurrences of each letter
char j = text[text.size()-1]; // sets j to the last character of text
if(j != '.' && j!= ' '&& j!=',' &&j!= '\0') //if the last character is not a period or empty space or comma or end of lone
words++; //add 1 to the count of words
cout<<"Number of words: "<<words<<endl; //display the number of words in the text
for(i=0; i<26; i++) { //iterates 25 times
if(letter[i]>0) //if letter at index i is greater than 0
cout<<(char)('a'+i)<<" : "<<letter[i]<<endl; }} //displays each letters and its number of occurrences in the text
Explanation:
The program is explained in the attached document with an example.
