Answer:
In Python:
num = int(input("Enter number: "))
maxn = num
while num >=0:
if num>maxn:
maxn = num
num = int(input("Enter number: "))
print("Largest: "+str(maxn))
Explanation:
Get input from the user
num = int(input("Enter number: "))
Initialize the largest to the first input
maxn = num
This loop is repeated until a negative input is recorded
while num >=0:
If the current input is greater than the previous largest
if num>maxn:
Set largest to the current input
maxn = num
Get another input from the user
num = int(input("Enter number: "))
Print the largest
print("Largest: "+str(maxn))
<span>Shared memory buffering would work best. This would give the ports the best allocation of resources, using only those that are the most active and best allocated for the size of the frames being transmitted in the current traffic. In addition, any port can store these frames, instead of being specifically allocated as per other types of memory buffering.</span>
Answer:
double ComputeGasVolume(double pressure, double temperature, double moles){
double volume = moles*GAS_CONST*temperature/pressure;
return volume;
}
Explanation:
You may insert this function just before your main function.
Create a function called ComputeGasVolume that takes three parameters, pressure, temperature, and moles
Using the given formula, PV = nRT, calculate the volume (V = nRT/P), and return it.
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
Answer:
Hi there! Pseudocode is the process of writing out the high-level structure of the program in simple English terms which acts as a blueprint of how the program will work. The pseudocode for this question is written below.
Explanation:
Prompt user for input 1
Validate input 1
Prompt user for input 2
Validate input 2
Prompt user for input 3
Validate input 3
Perform checks:
If
1 and 2 equals 3
Or
1 and 3 equals 2
Or
2 and 3 equals 1
Display message to user
