N 次抛硬币没有连续两个头像在一起的概率
给定一枚投掷 N 次的公平硬币，任务是确定没有两个头像连续出现的概率。
示例:
输入: N = 2 输出: 0.75 说明: 硬币掷 2 次，可能的结果是{TH，HT，TT，HH}。 因为在四分之三的结果中，头部不会同时出现。 因此，所需概率为(3/4)或 0.75
输入: N = 3 输出: 0.62 说明: 硬币掷 3 次，可能的结果是{TTT、HTT、THT、TTH、HHT、HTH、THH、HHH}。 因为八分之五的结果中，头部不会同时出现。 因此，所需概率为(5/8)或 0.62
方法:必须对有利结果的数量进行以下观察。
- N = 1 时:可能的结果是{T，H}。两者中有两个有利的结果。
- N = 2 时:可能的结果是{TH，HT，TT，HH}。四个中有三个有利结果。
- N = 3 时:同样，可能的结果是{TTT、HTT、THT、TTH、HHT、HTH、THH、HHH}。八个中有五个有利结果。
- N = 4 时:同样，可能的结果有{TTTT、TTTH、TTHT、THTT、HTTT、TTHH、THTH、HTHT、HHTT、THHT、HTTH、THHH、HTHH、HHTH、HHHT、hhhhhh }。十六个中有八个有利结果。
显然，有利结果的数量遵循斐波那契数列，其中 Fn(1) = 2，Fn(2) = 3，依此类推。因此，想法是实现斐波那契序列，以便找到有利案例的数量。显然，病例总数为 2 N 。 为了计算概率，使用以下公式:
P =有利案例/案例总数
下面是上述方法的实现:
C++
// C++ implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
#include <bits/stdc++.h>
using namespace std;
float round(float var,int digit)
{
float value = (int)(var *
pow(10, digit) + .5);
return (float)value /
pow(10, digit);
}
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
int probability(int N)
{
// The first two numbers in
// the sequence are initialized
int a = 2;
int b = 3;
// Base cases
if (N == 1)
{
return a;
}
else if(N == 2)
{
return b;
}
else
{
// Loop to compute the fibonacci
// sequence based on the first
// two initialized numbers
for(int i = 3; i <= N; i++)
{
int c = a + b;
a = b;
b = c;
}
return b;
}
}
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
float operations(int N)
{
// Computing the number of
// favourable cases
int x = probability(N);
// Computing the number of
// all possible outcomes for
// N tosses
int y = pow(2, N);
return round((float)x /
(float)y, 2);
}
// Driver code
int main()
{
int N = 10;
cout << (operations(N));
}
// Thus code is contributed by Rutvik_56
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
class GFG{
public static float round(float var, int digit)
{
float value = (int)(var *
Math.pow(10, digit) + .5);
return (float)value /
(float)Math.pow(10, digit);
}
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
public static int probability(int N)
{
// The first two numbers in
// the sequence are initialized
int a = 2;
int b = 3;
// Base cases
if (N == 1)
{
return a;
}
else if (N == 2)
{
return b;
}
else
{
// Loop to compute the fibonacci
// sequence based on the first
// two initialized numbers
for(int i = 3; i <= N; i++)
{
int c = a + b;
a = b;
b = c;
}
return b;
}
}
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
public static float operations(int N)
{
// Computing the number of
// favourable cases
int x = probability(N);
// Computing the number of
// all possible outcomes for
// N tosses
int y = (int)Math.pow(2, N);
return round((float)x /
(float)y, 2);
}
// Driver code
public static void main(String[] args)
{
int N = 10;
System.out.println((operations(N)));
}
}
// This code is contributed by divyeshrabadiya07
Python 3
# Python3 implementation to find the
# probability of not getting two
# consecutive heads together when
# N coins are tossed
import math
# Function to compute the N-th
# Fibonacci number in the
# sequence where a = 2
# and b = 3
def probability(N):
# The first two numbers in
# the sequence are initialized
a = 2
b = 3
# Base cases
if N == 1:
return a
elif N == 2:
return b
else:
# Loop to compute the fibonacci
# sequence based on the first
# two initialized numbers
for i in range(3, N + 1):
c = a + b
a = b
b = c
return b
# Function to find the probability
# of not getting two consecutive
# heads when N coins are tossed
def operations(N):
# Computing the number of
# favourable cases
x = probability (N)
# Computing the number of
# all possible outcomes for
# N tosses
y = math.pow(2, N)
return round(x / y, 2)
# Driver code
if __name__ == '__main__':
N = 10
print(operations(N))
C
// C# implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
using System;
class GFG{
public static float round(float var, int digit)
{
float value = (int)(var *
Math.Pow(10, digit) + .5);
return (float)value /
(float)Math.Pow(10, digit);
}
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
public static int probability(int N)
{
// The first two numbers in
// the sequence are initialized
int a = 2;
int b = 3;
// Base cases
if (N == 1)
{
return a;
}
else if (N == 2)
{
return b;
}
else
{
// Loop to compute the fibonacci
// sequence based on the first
// two initialized numbers
for(int i = 3; i <= N; i++)
{
int c = a + b;
a = b;
b = c;
}
return b;
}
}
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
public static float operations(int N)
{
// Computing the number of
// favourable cases
int x = probability(N);
// Computing the number of
// all possible outcomes for
// N tosses
int y = (int)Math.Pow(2, N);
return round((float)x /
(float)y, 2);
}
// Driver code
public static void Main(string[] args)
{
int N = 10;
Console.WriteLine((operations(N)));
}
}
// This code is contributed by chitranayal
java 描述语言
<script>
// javascript implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
function round(vr, digit) {
var value = parseInt( (vr* Math.pow(10, digit) + .5));
return value / Math.pow(10, digit);
}
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
function probability(N) {
// The first two numbers in
// the sequence are initialized
var a = 2;
var b = 3;
// Base cases
if (N == 1) {
return a;
} else if (N == 2) {
return b;
} else {
// Loop to compute the fibonacci
// sequence based on the first
// two initialized numbers
for (i = 3; i <= N; i++) {
var c = a + b;
a = b;
b = c;
}
return b;
}
}
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
function operations(N) {
// Computing the number of
// favourable cases
var x = probability(N);
// Computing the number of
// all possible outcomes for
// N tosses
var y = parseInt( Math.pow(2, N));
return round(x / y, 2);
}
// Driver code
var N = 10;
document.write((operations(N)));
// This code is contributed by aashish1995
</script>
Output:
0.14
时间复杂度:0(N)
辅助空间:0(1)
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