原文:
给定表示一个范围的两个整数 l 和 r ，任务是当在范围 l 到 r 中选择一个随机数时，找到得到一个数的。 举例:
输入: l = 6，r = 20 输出: 0.133333 解释: 范围内的完美方块【6，20】= { 9，16} = > 2 完美方块 范围内的总数【6，20】= 15 概率= 2 / 15 = 0.133333 输入: l = 16
方法:这个问题的关键观察是从 0 到一个数范围内的完美平方的计数可以用给定的公式计算:
//0 到 n 范围内的完美正方形的计数为 完美正方形的计数=地板(sqrt(n))
类似地，在给定范围内的完美正方形的计数可以借助于上述公式计算如下:
完美正方形的计数[l，r] =地板(sqrt(r))–天花板(sqrt(l)) 1 范围内的总数= r–l 1
以下是上述方法的实现:
c
// c implementation to find the
// probability of getting a
// perfect square number
#include
using namespace std;
// function to return the probability
// of getting a perfect square
// number in a range
float findprob(int l, int r)
{
// count of perfect squares
float countofps = floor(sqrt(r)) - ceil(sqrt(l)) 1;
// total numbers in range l to r
float total = r - l 1;
// calculating probability
float prob = (float)countofps / (float)total;
return prob;
}
// driver code
int main()
{
int l = 16, r = 25;
cout << findprob(l, r);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation to find the
// probability of getting a
// perfect square number
class gfg{
// function to return the probability
// of getting a perfect square
// number in a range
static float findprob(int l, int r)
{
// count of perfect squares
float countofps = (float) (math.floor(math.sqrt(r)) -
math.ceil(math.sqrt(l)) 1);
// total numbers in range l to r
float total = r - l 1;
// calculating probability
float prob = (float)countofps / (float)total;
return prob;
}
// driver code
public static void main(string[] args)
{
int l = 16, r = 25;
system.out.print(findprob(l, r));
}
}
// this code is contributed by amit katiyar
python 3
# python3 implementation to find
# the probability of getting a
# perfect square number
import math
# function to return the probability
# of getting a perfect square
# number in a range
def findprob(l, r):
# count of perfect squares
countofps = (math.floor(math.sqrt(r)) -
math.ceil(math.sqrt(l)) 1)
# total numbers in range l to r
total = r - l 1
# calculating probability
prob = countofps / total
return prob
# driver code
if __name__=='__main__':
l = 16
r = 25
print(findprob(l, r))
# this code is contributed by rutvik_56
c
// c# implementation to find the probability
// of getting a perfect square number
using system;
class gfg{
// function to return the probability
// of getting a perfect square
// number in a range
static float findprob(int l, int r)
{
// count of perfect squares
float countofps = (float)(math.floor(math.sqrt(r)) -
math.ceiling(math.sqrt(l)) 1);
// total numbers in range l to r
float total = r - l 1;
// calculating probability
float prob = (float)countofps / (float)total;
return prob;
}
// driver code
public static void main(string[] args)
{
int l = 16, r = 25;
console.write(findprob(l, r));
}
}
// this code is contributed by amit katiyar
java 描述语言
output:
0.2
时间复杂度: o(1)
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