原文:
给定两个数字 x 和 y ，任务是求10xt7】的任意正除数是10yt11】的整数倍的概率。 注: y 应为< = x. 例:****
输入: x = 2，y = 1 输出: 4/9 说明: 102的正因子为 1，2，4，5，10，20，25，50，100。一共 9 个。 101到 10 2 的倍数为 10，20，50，100。一共 4 个。 p(102的除数是 10 1 的倍数)= 4/9。 输入: x = 99，y = 88 输出: 9/625 解释:t30】a = 1099，b = 1088t35】p(10 的除数 99 是 10 的倍数 88 ) = 9/625
先决条件: 方法: 为了解决问题，我们需要观察以下步骤:
-
10 x 的所有除数将为:
(2p 5q*)，其中 0 < = p，q<= x
= t14】
-
求**10xt3 t5
10x= 2x* 5x
t14】的质因数数**
-
因此，10 x 的除数总数将为: ( x 1 ) * ( x 1 ) 。
-
现在，考虑 10 y 的所有倍数，它们可以是 10 x 的潜在约数。它们的形式还有:
(2a* 5b，其中 y < = a，b<= x
= t12】
-
所以10xt3【也是10yt7】的倍数的潜在约数为(x–y 1)(x–y 1)**。*
- 因此，所需概率为((x–y 1)(x–y 1))/((x 1)(x 1))。计算给定值 x 和 y 的表达式的值，给出了所需的概率。
以下是上述方法的实现:
c
// c program to find the probability
// of an arbitrary positive divisor of
// xth power of 10 to be a multiple of
// yth power of 10
#include
using namespace std;
#define int long long int
// function to calculate and print
// the required probability
void prob(int x, int y)
{
// count of potential divisors
// of x-th power of 10 which are
// also multiples of y-th power
// of 10
int num = abs(x - y 1)
* abs(x - y 1);
// count of divisors of x-th
// power of 10
int den = (x 1) * (x 1);
// calculate gcd
int gcd = __gcd(num, den);
// print the reduced
// fraction probability
cout << num / gcd << "/"
<< den / gcd << endl;
}
// driver code
signed main()
{
int x = 2, y = 1;
prob(x, y);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to find the probability
// of an arbitrary positive divisor of
// xth power of 10 to be a multiple of
// yth power of 10
import java.util.*;
class gfg{
// function to calculate and print
// the required probability
static void prob(int x, int y)
{
// count of potential divisors
// of x-th power of 10 which are
// also multiples of y-th power
// of 10
int num = math.abs(x - y 1) *
math.abs(x - y 1);
// count of divisors of x-th
// power of 10
int den = (x 1) * (x 1);
// calculate gcd
int gcd = __gcd(num, den);
// print the reduced
// fraction probability
system.out.print(num / gcd "/"
den / gcd "\n");
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// driver code
public static void main(string[] args)
{
int x = 2, y = 1;
prob(x, y);
}
}
// this code is contributed by amal kumar choubey
python 3
# python3 program to find the probability
# of an arbitrary positive divisor of
# xth power of 10 to be a multiple of
# yth power of 10
from math import *
# function to calculate and print
# the required probability
def prob(x, y):
# count of potential divisors
# of x-th power of 10 which are
# also multiples of y-th power
# of 10
num = abs(x - y 1) * abs(x - y 1)
# count of divisors of x-th
# power of 10
den = (x 1) * (x 1)
# calculate gcd
gcd1 = gcd(num, den)
# print the reduced
# fraction probability
print(num // gcd1, end = "")
print("/", end = "")
print(den // gcd1)
# driver code
if __name__ == '__main__':
x = 2
y = 1
prob(x, y)
# this code is contributed by surendra_gangwar
c
// c# program to find the probability
// of an arbitrary positive divisor of
// xth power of 10 to be a multiple of
// yth power of 10
using system;
class gfg{
// function to calculate and print
// the required probability
static void prob(int x, int y)
{
// count of potential divisors
// of x-th power of 10 which are
// also multiples of y-th power
// of 10
int num = math.abs(x - y 1) *
math.abs(x - y 1);
// count of divisors of x-th
// power of 10
int den = (x 1) * (x 1);
// calculate gcd
int gcd = __gcd(num, den);
// print the reduced
// fraction probability
console.write(num / gcd "/"
den / gcd "\n");
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// driver code
public static void main(string[] args)
{
int x = 2, y = 1;
prob(x, y);
}
}
// this code is contributed by ankitrai01
java 描述语言
output:
4/9
时间复杂度: o(log(n))
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