Answer:
There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:
Let a,b be two numbers.
c=a+b;
print(c);
But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:
Let a be the variable where the input numbers are stored.
while(f==1)
{
printf(“Enter number”);
scanf(“Take number into the variable a”);
sum=sum+a;
printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);
scanf(“Take the input into the variable f”);
}
print(Sum)
Explanation:
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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:
thanks for the knowledge! :D
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