Answer:
Ctrl+Space is the keyboard shortcut to select an entire column.
Explanation:
When you press the Shift+Space shortcut the first time it will select the entire row within the Table. Press Shift+Space a second time and it will select the entire row in the worksheet.
The same works for columns. Ctrl+Space will select the column of data in the Table. Pressing the keyboard shortcut a second time will include the column header of the Table in the selection. Pressing Ctrl+Space a third time will select the entire column in the worksheet.
You can select multiple rows or columns by holding Shift and pressing the Arrow Keys multiple times.
Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
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1) job postings - those offer include the phrases that the employers themselves want to hear
2) keyword lists - they're made for looking for keywords!
3) professional site's skills sections. You could use LinkedIn or any other professional site.
4)Sometimes you can find software that can suggest keywords for you
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;
}
