一次 2 步或 3 步到达某点的概率
一个人从 X = 0 的位置开始走，如果她只能走 2 步或 3 步，找到恰好到达 X = N 的概率。给出步长 2 的概率，即步长 3 的概率为 1–P 例:
Input : N = 5, P = 0.20
Output : 0.32
Explanation :-
There are two ways to reach 5.
2+3 with probability = 0.2 * 0.8 = 0.16
3+2 with probability = 0.8 * 0.2 = 0.16
So, total probability = 0.32.
这是一个简单的动态规划问题。它是这个问题的简单延伸:- 计数-不同方式-表达-n-和-1-3-4 下面是上述方法的实现。
C++
// CPP Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
#include <bits/stdc++.h>
using namespace std;
// Returns probability to reach N
float find_prob(int N, float P)
{
double dp[N + 1];
dp[0] = 1;
dp[1] = 0;
dp[2] = P;
dp[3] = 1 - P;
for (int i = 4; i <= N; ++i)
dp[i] = (P)*dp[i - 2] + (1 - P) * dp[i - 3];
return dp[N];
}
// Driver code
int main()
{
int n = 5;
float p = 0.2;
cout << find_prob(n, p);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
import java.io.*;
class GFG {
// Returns probability to reach N
static float find_prob(int N, float P)
{
double dp[] = new double[N + 1];
dp[0] = 1;
dp[1] = 0;
dp[2] = P;
dp[3] = 1 - P;
for (int i = 4; i <= N; ++i)
dp[i] = (P) * dp[i - 2] +
(1 - P) * dp[i - 3];
return ((float)(dp[N]));
}
// Driver code
public static void main(String args[])
{
int n = 5;
float p = 0.2f;
System.out.printf("%.2f",find_prob(n, p));
}
}
/* This code is contributed by Nikita Tiwari.*/
Python 3
# Python 3 Program to find
# probability to reach N with
# P probability to take 2
# steps (1-P) to take 3 steps
# Returns probability to reach N
def find_prob(N, P) :
dp =[0] * (n + 1)
dp[0] = 1
dp[1] = 0
dp[2] = P
dp[3] = 1 - P
for i in range(4, N + 1) :
dp[i] = (P) * dp[i - 2] + (1 - P) * dp[i - 3]
return dp[N]
# Driver code
n = 5
p = 0.2
print(round(find_prob(n, p), 2))
# This code is contributed by Nikita Tiwari.
C
// C# Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
using System;
class GFG {
// Returns probability to reach N
static float find_prob(int N, float P)
{
double []dp = new double[N + 1];
dp[0] = 1;
dp[1] = 0;
dp[2] = P;
dp[3] = 1 - P;
for (int i = 4; i <= N; ++i)
dp[i] = (P) * dp[i - 2] +
(1 - P) * dp[i - 3];
return ((float)(dp[N]));
}
// Driver code
public static void Main()
{
int n = 5;
float p = 0.2f;
Console.WriteLine(find_prob(n, p));
}
}
/* This code is contributed by vt_m.*/
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
// Returns probability to reach N
function find_prob($N, $P)
{
$dp;
$dp[0] = 1;
$dp[1] = 0;
$dp[2] = $P;
$dp[3] = 1 - $P;
for ($i = 4; $i <= $N; ++$i)
$dp[$i] = ($P) * $dp[$i - 2] +
(1 - $P) * $dp[$i - 3];
return $dp[$N];
}
// Driver code
$n = 5;
$p = 0.2;
echo find_prob($n, $p);
// This code is contributed by mits.
?>
java 描述语言
<script>
// JavaScript Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
// Returns probability to reach N
function find_prob(N, P)
{
let dp = [];
dp[0] = 1;
dp[1] = 0;
dp[2] = P;
dp[3] = 1 - P;
for (let i = 4; i <= N; ++i)
dp[i] = (P) * dp[i - 2] +
(1 - P) * dp[i - 3];
return (dp[N]);
}
// Driver Code
let n = 5;
let p = 0.2;
document.write(find_prob(n, p));
// This code is contributed by chinmoy1997pal.
</script>
输出:
0.32
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