当给定单个击中目标的概率时，A 赢得比赛的概率
给定四个整数 a 、 b 、 c 和 d 。玩家 A & B 尝试点球得分。A 射击目标的概率为 a / b ，B 射击目标的概率为 c / d 。先罚点球的球员获胜。任务是找出 A 赢得比赛的概率。 举例:
输入: a = 1，b = 3，c = 1，d = 3 输出: 0.6 输入: a = 1，b = 2，c = 10，d = 11 输出: 0.52381
进场:如果我们考虑变量 K = a / b 为 A 击中目标的概率，R =(1 –( A/B))*(1–(c/d))为 A 和 B 都没有击中目标的概率。 因此，解形成几何级数K * R0+K * R1+K * R2+…..其和为(K/1–R)。将 K 和 R 的值放在一起后，我们得到公式为K (1/(1 –( 1–R)(1–K))。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the probability of A winning
double getProbability(int a, int b, int c, int d)
{
// p and q store the values
// of fractions a / b and c / d
double p = (double)a / (double)b;
double q = (double)c / (double)d;
// To store the winning probability of A
double ans = p * (1 / (1 - (1 - q) * (1 - p)));
return ans;
}
// Driver code
int main()
{
int a = 1, b = 2, c = 10, d = 11;
cout << getProbability(a, b, c, d);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the probability
// of A winning
static double getProbability(int a, int b,
int c, int d)
{
// p and q store the values
// of fractions a / b and c / d
double p = (double) a / (double) b;
double q = (double) c / (double) d;
// To store the winning probability of A
double ans = p * (1 / (1 - (1 - q) *
(1 - p)));
return ans;
}
// Driver code
public static void main(String[] args)
{
int a = 1, b = 2, c = 10, d = 11;
System.out.printf("%.5f",
getProbability(a, b, c, d));
}
}
// This code contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function to return the probability
# of A winning
def getProbability(a, b, c, d) :
# p and q store the values
# of fractions a / b and c / d
p = a / b;
q = c / d;
# To store the winning probability of A
ans = p * (1 / (1 - (1 - q) * (1 - p)));
return round(ans,5);
# Driver code
if __name__ == "__main__" :
a = 1; b = 2; c = 10; d = 11;
print(getProbability(a, b, c, d));
# This code is contributed by Ryuga
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the probability
// of A winning
public static double getProbability(int a, int b,
int c, int d)
{
// p and q store the values
// of fractions a / b and c / d
double p = (double) a / (double) b;
double q = (double) c / (double) d;
// To store the winning probability of A
double ans = p * (1 / (1 - (1 - q) *
(1 - p)));
return ans;
}
// Driver code
public static void Main(string[] args)
{
int a = 1, b = 2, c = 10, d = 11;
Console.Write("{0:F5}",
getProbability(a, b, c, d));
}
}
// This code is contributed by Shrikant13
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the probability
// of A winning
function getProbability($a, $b, $c, $d)
{
// p and q store the values
// of fractions a / b and c / d
$p = $a / $b;
$q = $c / $d;
// To store the winning probability of A
$ans = $p * (1 / (1 - (1 - $q) * (1 - $p)));
return round($ans,6);
}
// Driver code
$a = 1;
$b = 2;
$c = 10;
$d = 11;
echo getProbability($a, $b, $c, $d);
// This code is contributed by chandan_jnu
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the probability
// of A winning
function getProbability(a , b , c , d) {
// p and q store the values
// of fractions a / b and c / d
var p = a / b;
var q = c / d;
// To store the winning probability of A
var ans = p * (1 / (1 - (1 - q) * (1 - p)));
return ans;
}
// Driver code
var a = 1, b = 2, c = 10, d = 11;
document.write( getProbability(a, b, c, d).toFixed(5));
// This code contributed by aashish1995
</script>
Output:
0.52381
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