原文:
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
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）
java 描述语言
output:
0.52381
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