Taking the input as a string, slicing it with a space as the delimiter, turning the divided portion into an integer, and then appending to the list.
<h3>
Explanation:</h3>
def selection_sort_descend_trace(numbers):
i=len(numbers)
print("Output: ")
for num in range(0,i-1):# traversing from 0 to N-2, total N-1 iterations
val=numbers[num]
start=num+1
end=i
t=0
for j in range(start,end):
if(val<numbers[j]):
remember=j
val=numbers[j]
t=1
if(t==1):# swaping onlf if greater number is available
temp=numbers[num]
numbers[num]=val
numbers[remember]=temp
for p in range(i):# printing
print(numbers[p],end=' ')
print("\n")
if __name__ == "__main__":
print("Enter the integers separated by space: ")
numbers=[int(x) for x in input().split(' ')]
selection_sort_descend_trace(numbers)
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Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
<u></u>
Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
Answer:
Direct Mapped Cache
Explanation:
Given that a Direct Mapped Cache is a form of mapping whereby each main memory address is mapped into precisely one cache block.
It is considered cheaper compared to the associative method of cache mapping, and it is faster when searching through it. This is because it utilizes a tag field only.
Hence, The method of mapping where each memory location is mapped to exactly one location in the cache is "Direct Mapped Cache"
Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
int sum = 0, n;
do {cin>>n; sum+=n;}while (n!=0);
cout<<sum;
