原文:
给定三个整数 l 、 r 和 m ，任务是求【l，r】范围内一个数的被 m 整除的概率
欧拉全能性函数是{1，2，3，…，n}中相对于 n 为素数的数的计数，即 gcd(最大公约数)与 n 为 1 的数。
示例:
输入: l = 1，r = 5，m = 2 输出: 0.6 说明: n = 1，2，3，4，5 的欧拉全能性函数分别为{1，1，2，2，4}。 可被 m(= 2)整除的欧拉全能函数的计数为 3。 因此，所需概率为 3/5 = 0.6
输入: l = 1，r = 7，m = 4 输出: 0.142 解释: n = 1，2，3，…7 的欧拉全能性函数分别为{1，1，2，2，4，2，6}。 可被 m(= 4)整除的欧拉全能性函数的计数为 1。 因此，所需概率为 1/7 = 0.142
方法:思路是预先计算，在给定范围内迭代，计算可被 m 整除的数，计算概率。
对于的计算，使用欧拉乘积公式:
其中pit3】是 n 的质因数。
对于*(l<= n<= r)*的每个质因数 i ，执行以下步骤:
- 从【1，n】中减去 i 的所有倍数。
- 通过重复除以 i 来更新 n 。
- 如果减少的 n 大于 1 ，则从结果中去除所有倍数的 n 。
质因数的计算采用厄拉多塞法的筛。给定范围内的概率为计数/(l–r 1)。
下面是上述方法的实现:
c
// c program to implement
// the above approach
#include
using namespace std;
#define size 1000001
// seieve of erotosthenes
// to compute all primes
void seiveoferatosthenes(int* prime)
{
prime[0] = 1, prime[1] = 0;
for (int i = 2; i * i < 1000001; i ) {
// if prime
if (prime[i] == 0) {
for (int j = i * i; j < 1000001;
j = i) {
// mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// function to find the probability of
// euler's totient function in a given range
float probabiltyeuler(int* prime, int l,
int r, int m)
{
int* arr = new int[size]{ 0 };
int* eulertotient = new int[size]{ 0 };
int count = 0;
// initializing two arrays
// with values from l to r
// for euler's totient
for (int i = l; i <= r; i ) {
// indexing from 0
eulertotient[i - l] = i;
arr[i - l] = i;
}
for (int i = 2; i < 1000001; i ) {
// if the current number is prime
if (prime[i] == 0) {
// checking if i is prime factor
// of numbers in range l to r
for (int j = (l / i) * i; j <= r;
j = i) {
if (j - l >= 0) {
// update all the numbers
// which has prime factor i
eulertotient[j - l]
= eulertotient[j - l]
/ i * (i - 1);
while (arr[j - l] % i == 0) {
arr[j - l] /= i;
}
}
}
}
}
// if number in range has a
// prime factor > sqrt(number)
for (int i = l; i <= r; i ) {
if (arr[i - l] > 1) {
eulertotient[i - l]
= (eulertotient[i - l] / arr[i - l])
* (arr[i - l] - 1);
}
}
for (int i = l; i <= r; i ) {
// count those which are divisible by m
if ((eulertotient[i - l] % m) == 0) {
count ;
}
}
// return the result
return (1.0 * count / (r 1 - l));
}
// driver code
int main()
{
int* prime = new int[size]{ 0 };
seiveoferatosthenes(prime);
int l = 1, r = 7, m = 3;
cout << probabiltyeuler(prime, l, r, m);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to implement
// the above approach
import java.util.*;
class gfg{
static final int size = 1000001;
// seieve of erotosthenes
// to compute all primes
static void seiveoferatosthenes(int []prime)
{
prime[0] = 1;
prime[1] = 0;
for (int i = 2; i * i < 1000001; i )
{
// if prime
if (prime[i] == 0)
{
for (int j = i * i; j < 1000001; j = i)
{
// mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// function to find the probability of
// euler's totient function in a given range
static float probabiltyeuler(int []prime, int l,
int r, int m)
{
int[] arr = new int[size];
int []eulertotient = new int[size];
int count = 0;
// initializing two arrays
// with values from l to r
// for euler's totient
for (int i = l; i <= r; i )
{
// indexing from 0
eulertotient[i - l] = i;
arr[i - l] = i;
}
for (int i = 2; i < 1000001; i )
{
// if the current number is prime
if (prime[i] == 0)
{
// checking if i is prime factor
// of numbers in range l to r
for (int j = (l / i) * i; j <= r; j = i)
{
if (j - l >= 0)
{
// update all the numbers
// which has prime factor i
eulertotient[j - l] = eulertotient[j - l] /
i * (i - 1);
while (arr[j - l] % i == 0)
{
arr[j - l] /= i;
}
}
}
}
}
// if number in range has a
// prime factor > math.sqrt(number)
for (int i = l; i <= r; i )
{
if (arr[i - l] > 1)
{
eulertotient[i - l] = (eulertotient[i - l] / arr[i - l]) *
(arr[i - l] - 1);
}
}
for (int i = l; i <= r; i )
{
// count those which are divisible by m
if ((eulertotient[i - l] % m) == 0)
{
count ;
}
}
// return the result
return (float) (1.0 * count / (r 1 - l));
}
// driver code
public static void main(string[] args)
{
int []prime = new int[size];
seiveoferatosthenes(prime);
int l = 1, r = 7, m = 3;
system.out.print(probabiltyeuler(prime, l, r, m));
}
}
// this code is contributed by sapnasingh4991
python 3
# python3 program to implement
# the above approach
size = 1000001
# seieve of erotosthenes
# to compute all primes
def seiveoferatosthenes(prime):
prime[0] = 1
prime[1] = 0
i = 2
while(i * i < 1000001):
# if prime
if (prime[i] == 0):
j = i * i
while(j < 1000001):
# mark all its multiples
# as non-prime
prime[j] = 1
j = j i
i = 1
# function to find the probability of
# euler's totient function in a given range
def probabiltyeuler(prime, l, r, m):
arr = [0] * size
eulertotient = [0] * size
count = 0
# initializing two arrays
# with values from l to r
# for euler's totient
for i in range(l, r 1):
# indexing from 0
eulertotient[i - l] = i
arr[i - l] = i
for i in range(2, 1000001):
# if the current number is prime
if (prime[i] == 0):
# checking if i is prime factor
# of numbers in range l to r
for j in range((l // i) * i, r 1, i):
if (j - l >= 0):
# update all the numbers
# which has prime factor i
eulertotient[j - l] = (eulertotient[j - l] //
i * (i - 1))
while (arr[j - l] % i == 0):
arr[j - l] = arr[j - l] // i
# if number in range has a
# prime factor > math.sqrt(number)
for i in range(l, r 1):
if (arr[i - l] > 1):
eulertotient[i - l] = ((eulertotient[i - l] //
arr[i - l]) *
(arr[i - l] - 1))
for i in range(l, r 1):
# count those which are divisible by m
if ((eulertotient[i - l] % m) == 0):
count = 1
# return the result
return (float)(1.0 * count / (r 1 - l))
# driver code
prime = [0] * size
seiveoferatosthenes(prime)
l, r, m = 1, 7, 3
print(probabiltyeuler(prime, l, r, m))
# this code is contributed by divyeshrabadiya07
c#
// c# program to implement
// the above approach
using system;
class gfg{
static readonly int size = 1000001;
// seieve of erotosthenes
// to compute all primes
static void seiveoferatosthenes(int []prime)
{
prime[0] = 1;
prime[1] = 0;
for (int i = 2; i * i < 1000001; i )
{
// if prime
if (prime[i] == 0)
{
for (int j = i * i; j < 1000001; j = i)
{
// mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// function to find the probability of
// euler's totient function in a given range
static float probabiltyeuler(int []prime, int l,
int r, int m)
{
int[] arr = new int[size];
int []eulertotient = new int[size];
int count = 0;
// initializing two arrays
// with values from l to r
// for euler's totient
for (int i = l; i <= r; i )
{
// indexing from 0
eulertotient[i - l] = i;
arr[i - l] = i;
}
for (int i = 2; i < 1000001; i )
{
// if the current number is prime
if (prime[i] == 0)
{
// checking if i is prime factor
// of numbers in range l to r
for (int j = (l / i) * i; j <= r; j = i)
{
if (j - l >= 0)
{
// update all the numbers
// which has prime factor i
eulertotient[j - l] = eulertotient[j - l] /
i * (i - 1);
while (arr[j - l] % i == 0)
{
arr[j - l] /= i;
}
}
}
}
}
// if number in range has a
// prime factor > math.sqrt(number)
for (int i = l; i <= r; i )
{
if (arr[i - l] > 1)
{
eulertotient[i - l] = (eulertotient[i - l] / arr[i - l]) *
(arr[i - l] - 1);
}
}
for (int i = l; i <= r; i )
{
// count those which are divisible by m
if ((eulertotient[i - l] % m) == 0)
{
count ;
}
}
// return the result
return (float) (1.0 * count / (r 1 - l));
}
// driver code
public static void main(string[] args)
{
int []prime = new int[size];
seiveoferatosthenes(prime);
int l = 1, r = 7, m = 3;
console.write(probabiltyeuler(prime, l, r, m));
}
}
// this code is contributed by sapnasingh4991
java 描述语言
**output:
0.142857**
*时间复杂度: o(nlog(n))辅助空间: o(大小)，其中大小表示计算筛的个数。***
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