给定一个由1和0组成的数组,其中A i = 1表示第i天是一个雨天,而A i = 0表示这不是一个雨天。任务是找到的概率N + 1次是一个雨天。
例子:
Input: a[] = {0, 0, 1, 0}
Output: .25
Since one day was rainy out of 4 days, hence the probability on
5th day will be 0.25
Input: a[] = {1, 0, 1, 0, 1, 1, 1}
Output: 0.71
可以使用以下公式找出第N + 1天下雨的概率:
Probability = number of rainy days / total number of days.
首先,计算1的个数,然后将概率乘以1的个数除以N,即count /N。
下面是上述方法的实现:
C++
// C++ code to find the probability of rain
// on n+1-th day when previous day's data is given
#include
using namespace std;
// Function to find the probability
float rainDayProbability(int a[], int n)
{
float count = 0, m;
// count 1
for (int i = 0; i < n; i++) {
if (a[i] == 1)
count++;
}
// find probability
m = count / n;
return m;
}
// Driver Code
int main()
{
int a[] = { 1, 0, 1, 0, 1, 1, 1, 1 };
int n = sizeof(a) / sizeof(a[0]);
cout << rainDayProbability(a, n);
return 0;
}
Java
// Java code to find the
// probability of rain
// on n+1-th day when previous
// day's data is given
import java.io.*;
import java.util.*;
class GFG
{
// Function to find
// the probability
static float rainDayProbability(int a[],
int n)
{
float count = 0, m;
// count 1
for (int i = 0; i < n; i++)
{
if (a[i] == 1)
count++;
}
// find probability
m = count / n;
return m;
}
// Driver Code
public static void main(String args[])
{
int a[] = { 1, 0, 1, 0, 1, 1, 1, 1 };
int n = a.length;
System.out.print(rainDayProbability(a, n));
}
}
Python 3
# Python 3 program to find
# the probability of rain
# on n+1-th day when previous
# day's data is given
# Function to find the probability
def rainDayProbability(a, n) :
# count occurence of 1
count = a.count(1)
# find probability
m = count / n
return m
# Driver code
if __name__ == "__main__" :
a = [ 1, 0, 1, 0, 1, 1, 1, 1]
n = len(a)
# function calling
print(rainDayProbability(a, n))
# This code is contributed
# by ANKITRAI1
C#
// C# code to find the
// probability of rain
// on n+1-th day when
// previous day's data
// is given
using System;
class GFG
{
// Function to find
// the probability
static float rainDayProbability(int []a,
int n)
{
float count = 0, m;
// count 1
for (int i = 0; i < n; i++)
{
if (a[i] == 1)
count++;
}
// find probability
m = count / n;
return m;
}
// Driver Code
public static void Main()
{
int []a = {1, 0, 1, 0,
1, 1, 1, 1};
int n = a.Length;
Console.WriteLine(rainDayProbability(a, n));
}
}
// This code is contributed
// by inder_verma.
PHP
Javascript
输出:
0.75
时间复杂度: O(N)