原文:
给定一个包含 n 元素的集合。如果两个子集 x 和 y 被选中，那么找到它们都包含相同数量元素的概率。 举例:
输入:4 t3】输出: 35/128 输入: 2 输出: 3/8
方法: 我们选择一个子集 x 有 r 个元素，那么 y 必须包含 r 个元素。一个子集可以有最小 0 元素和最大 n 元素。 一个集合包含的子集总数 n 元素个数为
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，同时选择 x 和 y 的总可能方式将是
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=
*** quicklatex cannot compile formula:
*** error message:
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=
*** quicklatex cannot compile formula:
*** error message:
error: nothing to show, formula is empty
。 let， p =选择 x 和 y 的总可能方式，使两者具有相同的元素数量。 那么 p = = = 那么要求的概率就是。 以下是上述方法的实施:
c
// c implementation of
// the above approach
#include
using namespace std;
// returns value of binomial
// coefficient c(n, k)
int binomialcoeff(int n, int k)
{
int res = 1;
// since c(n, k) = c(n, n-k)
if (k > n - k)
k = n - k;
// calculate value of
for (int i = 0; i < k; i) {
res *= (n - i);
res /= (i 1);
}
return res;
}
// iterative function to
// calculate (x^y) in o(log y)
int power(int x, unsigned int y)
{
// initialize result
int res = 1;
while (y > 0) {
// if y is odd, multiply
// x with result
if (y & 1)
res = res * x;
// y must be even now
// y = y/2
y = y >> 1;
// change x to x^2
x = x * x;
}
return res;
}
// function to find probability
void findprobability(int n)
{
// calculate total possible
// ways and favourable ways.
int up = binomialcoeff(2 * n, n);
int down = power(2, 2 * n);
// divide by gcd such that
// they become relatively coprime
int g = __gcd(up, down);
up /= g, down /= g;
cout << up << "/" << down << endl;
}
// driver code
int main()
{
int n = 8;
findprobability(n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of
// the above approach
class gfg
{
// returns value of binomial
// coefficient c(n, k)
static int binomialcoeff(int n, int k)
{
int res = 1;
// since c(n, k) = c(n, n-k)
if (k > n - k)
k = n - k;
// calculate value of
for (int i = 0; i < k; i)
{
res *= (n - i);
res /= (i 1);
}
return res;
}
// iterative function to
// calculate (x^y) in o(log y)
static int power(int x, int y)
{
// initialize result
int res = 1;
while (y > 0)
{
// if y is odd, multiply
// x with result
if ((y & 1) == 1)
res = res * x;
// y must be even now
// y = y/2
y = y >> 1;
// change x to x^2
x = x * x;
}
return res;
}
// recursive function to return gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// function to find probability
static void findprobability(int n)
{
// calculate total possible
// ways and favourable ways.
int up = binomialcoeff(2 * n, n);
int down = power(2, 2 * n);
// divide by gcd such that
// they become relatively coprime
int g = gcd(up, down);
up /= g;
down /= g;
system.out.println(up "/" down);
}
// driver code
public static void main (string[] args)
{
int n = 8;
findprobability(n);
}
}
// this code is contributed by ankitrai01
python 3
# python3 implementation of
# the above approach
import math
# returns value of binomial
# coefficient c(n, k)
def binomialcoeff(n, k):
res = 1
# since c(n, k) = c(n, n-k)
if (k > n - k):
k = n - k
# calculate value of
for i in range(0, k):
res = res * (n - i)
res = res // (i 1)
return res
# iterative function to
# calculate (x^y) in o(log y)
def power(x, y):
# initialize result
res = 1
while (y > 0):
# if y is odd, multiply
# x with result
if (y & 1):
res = res * x
# y must be even now
# y = y/2
y = y // 2
# change x to x^2
x = x * x
return res
# function to find probability
def findprobability(n):
# calculate total possible
# ways and favourable ways.
up = binomialcoeff(2 * n, n)
down = power(2, 2 * n)
# divide by gcd such that
# they become relatively coprime
g = math.gcd(up,down)
up = up // g
down = down // g
print(up, "/", down)
# driver code
n = 8
findprobability(n)
# this code is contributed by sanjit_prasad
c
// c# implementation of
// the above approach
using system;
using system.collections.generic;
class gfg
{
// returns value of binomial
// coefficient c(n, k)
static int binomialcoeff(int n, int k)
{
int res = 1;
// since c(n, k) = c(n, n-k)
if (k > n - k)
k = n - k;
// calculate value of
for (int i = 0; i < k; i)
{
res *= (n - i);
res /= (i 1);
}
return res;
}
// iterative function to
// calculate (x^y) in o(log y)
static int power(int x, int y)
{
// initialize result
int res = 1;
while (y > 0)
{
// if y is odd, multiply
// x with result
if ((y & 1) == 1)
res = res * x;
// y must be even now
// y = y/2
y = y >> 1;
// change x to x^2
x = x * x;
}
return res;
}
// recursive function to
// return gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// function to find probability
static void findprobability(int n)
{
// calculate total possible
// ways and favourable ways.
int up = binomialcoeff(2 * n, n);
int down = power(2, 2 * n);
// divide by gcd such that
// they become relatively coprime
int g = gcd(up, down);
up /= g;
down /= g;
console.writeline(up "/" down);
}
// driver code
public static void main (string[] args)
{
int n = 8;
findprobability(n);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
6435/32768
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