给定字符C和一个整数N，表示C位置的N个硬币，其中C可以是头或尾。我们可以翻转硬币N次，其中第i轮玩家将翻转所有小于或等于i的硬币的面孔。任务是确定翻转N次可能的时间后头和尾的总数。
例子：
Input: C = ‘H’, N = 5
Output: Head = 2, Tail = 3
Explanation:
H means initially all the coins are facing in head direction, N means the total number of coins.
So initially for i = 0, we have: H H H H H
After the first round that is i = 1: T H H H H
After the second round that is i = 2: H T H H H
After the third round that is i = 3: T H T H H
After the fourth round that is i = 4: H T H T H
After the fifth round that is i = 5: T H T H T
Hence the total count of the head is 2 and tail is 3.
Input: C = ‘T’, N = 7
Output: Head = 4, Tail = 3
Explanation:
After all the possible flips the head and tail count is 4 and 3.
方法：
为了解决上述问题，我们必须遵循以下步骤：
- 在上面的问题中，如果我们观察到，则存在一种模式，如果最初所有硬币都朝向头方向，则N轮后的头总数将为(n / 2)的底值，而尾数将为单元格(n / 2)的值。
- 否则，如果所有硬币都朝向尾巴方向，则N轮后的尾巴总数将为底值(n / 2)，而头将为ceil值(n / 2)。
下面是实现：
C++
// C++ program to count total heads
// and tails after N flips in a coin
#include
using namespace std;
// Function to find count of head and tail
pair count_ht(char s, int N)
{
// Check if initially all the
// coins are facing towards head
pairp;
if(s == 'H')
{
p.first = floor(N / 2.0);
p.second = ceil(N / 2.0);
}
// Check if initially all the coins
// are facing towards tail
else if(s == 'T')
{
p.first = ceil(N / 2.0);
p.second = floor(N / 2.0);
}
return p;
}
// Driver code
int main()
{
char C = 'H';
int N = 5;
pair p = count_ht(C, N);
cout << "Head = " << (p.first) << "\n";
cout << "Tail = " << (p.second) << "\n";
}
// This code is contributed by virusbuddah_ Java
// Java program to count
// total heads and tails
// after N flips in a coin
import javafx.util.Pair;
public class Main
{
// Function to find count of head and tail
public static Pair count_ht(char s,
int N)
{
// Check if initially all the
// coins are facing towards head
Pair p = new Pair (0, 0);
if(s == 'H')
{
p = new Pair ((int)Math.floor(N / 2.0),
(int)Math.ceil(N / 2.0));
}
// Check if initially all the coins
// are facing towards tail
else if(s == 'T')
{
p = new Pair ((int)Math.ceil(N / 2.0),
(int)Math.floor(N / 2.0));
}
return p;
}
public static void main(String[] args)
{
char C = 'H';
int N = 5;
Pair p = count_ht(C, N);
System.out.println("Head = " + p.getKey());
System.out.println("Tail = " + p.getValue());
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 program to Count total heads
# and tails after N flips in a coin
# Function to find count of head and tail
import math
def count_ht( s, N ):
# Check if initially all the
# coins are facing towards head
if s == "H":
h = math.floor( N / 2 )
t = math.ceil( N / 2 )
# Check if initially all the coins
# are facing towards tail
elif s == "T":
h = math.ceil( N / 2 )
t = math.floor( N / 2 )
return [h, t]
# Driver Code
if __name__ == "__main__":
C = "H"
N = 5
l = count_ht(C, n)
print("Head = ", l[0])
print("Tail = ", l[1])C#
// C# program to count total heads
// and tails after N flips in a coin
using System;
class GFG{
// Function to find count of head and tail
public static Tuple count_ht(char s,
int N)
{
// Check if initially all the
// coins are facing towards head
Tuple p = Tuple.Create(0, 0);
if (s == 'H')
{
p = Tuple.Create((int)Math.Floor(N / 2.0),
(int)Math.Ceiling(N / 2.0));
}
// Check if initially all the coins
// are facing towards tail
else if (s == 'T')
{
p = Tuple.Create((int)Math.Ceiling(N / 2.0),
(int)Math.Floor(N / 2.0));
}
return p;
}
// Driver Code
static void Main()
{
char C = 'H';
int N = 5;
Tuple p = count_ht(C, N);
Console.WriteLine("Head = " + p.Item1);
Console.WriteLine("Tail = " + p.Item2);
}
}
// This code is contributed by divyesh072019 Head = 2
Tail = 3时间复杂度： O(1)