原文:
给定一个数组 a[]，数组的大小是 n，另一个是 key k，任务是找出 key k 出现在数组中的概率。 例:
input : n = 6
a[] = { 4, 7, 2, 0, 8, 7, 5 }
k = 3
output :0
since value of k = 3 is not present in array,
hence the probability of 0.
input :n = 10
a[] = { 2, 3, 5, 1, 9, 8, 0, 7, 6, 5 }
k = 5
output :0.2
的概率可以用下面的公式求出:
probability = total number of k present /
size of array.
首先计算 k 的数量，然后概率将是 k 的数量除以 n，即 count/n 下面是上述方法的实现:
c
// c code to find the probability of
// search key k present in array
#include
using namespace std;
// function to find the probability
float kpresentprobability(int a[], int n, int k)
{
float count = 0;
for (int i = 0; i < n; i )
if (a[i] == k)
count ;
// find probability
return count / n;
}
// driver code
int main()
{
int a[] = { 4, 7, 2, 0, 8, 7, 5 };
int k = 3;
int n = sizeof(a) / sizeof(a[0]);
cout << kpresentprobability(a, n, k);
return 0;
}
python 3
# python3 code to find the
# probability of search key
# k present in 1d-array (list).
# function to find the probability
def kpresentprobability(a, n, k) :
count = a.count(k)
# find probability upto
# 2 decimal places
return round(count / n , 2)
# driver code
if __name__ == "__main__" :
a = [ 4, 7, 2, 0, 8, 7, 5 ]
k = 2
n = len(a)
print(kpresentprobability( a, n, k))
# this code is contributed
# by ankitrai1
java 语言(一种计算机语言，尤用于创建网站)
// java code to find the probability
// of search key k present in array
class gfg
{
// function to find the probability
static float kpresentprobability(int a[],
int n,
int k)
{
float count = 0;
for (int i = 0; i < n; i )
if (a[i] == k)
count ;
// find probability
return count/ n;
}
// driver code
public static void main(string[] args)
{
int a[] = { 4, 7, 2, 0, 8, 7, 5 };
int k = 2;
int n = a.length;
double n = kpresentprobability(a, n, k);
double p = (double)math.round(n * 100) / 100;
system.out.println(p);
}
}
// this code is contributed
// by chitranayal
c
// c# code to find the probability
// of search key k present in array
using system;
class gfg
{
// function to find the probability
static float kpresentprobability(int[] a,
int n,
int k)
{
float count = 0;
for (int i = 0; i < n; i )
if (a[i] == k)
count ;
// find probability
return count/ n;
}
// driver code
public static void main()
{
int[] a = { 4, 7, 2, 0, 8, 7, 5 };
int k = 2;
int n = a.length;
double n = kpresentprobability(a, n, k);
double p = (double)math.round(n * 100) / 100;
console.write(p);
}
}
// this code is contributed
// by chitranayal
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
output
0
时间复杂度:o(n)
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