给定两个数组arr1 []和arr2 [] ，玩家A从arr1 []中选择一个元素，玩家B从arr2 []中选择一个元素，该元素具有除数更多的元素的玩家将赢得本轮比赛。如果两个元素的除数均相同，则玩家A赢得该回合。任务是确定玩家A或玩家B赢得游戏的可能性更大。
例子：
Input: arr1[] = {4, 12, 24}, arr2[] = {25, 28, 13, 45}
Output: A
Pairs where A wins are (3, 2), (3, 3), (6, 2), (6, 3), (6, 6), (6, 6), (8, 2), (8, 3), (8, 6) and (8, 6).
Total = 10.
B wins in 2 cases.
Hence, A has more probability of winning the game.
Input: arr1[] = {7, 3, 4}, arr2[] = {5, 4, 12, 10}
Output: B
方法：
- 对于这两个数组，代替元素，存储元素的除数的数量。
- 将两个数组按升序排序。
- 找到A赢得游戏的所有可能的对选择(X，Y) 。
- 假设A从arr1 []中选择一个元素。现在，可以使用二进制搜索来查找arr2 []中的元素数，该元素的除数小于arr1 []中的所选元素。这将被添加到A获胜的对数中。对arr1 []的每个元素都执行此操作。
- N * M是可能的总数对。 A获胜的对数为X ， B获胜的对数为Y。
- 如果X> Y，则A更有可能赢得比赛。
- 如果Y> X，则B具有更大的获胜概率。
- 如果X = Y，则抽奖的可能性更大。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long int
// Function to return the count
// of divisors of elem
int divisorcount(int elem)
{
int ans = 0;
for (int i = 1; i <= sqrt(elem); i++) {
if (elem % i == 0) {
if (i * i == elem)
ans++;
else
ans += 2;
}
}
return ans;
}
// Function to return the winner of the game
string findwinner(int A[], int B[], int N, int M)
{
// Convert every element of A[]
// to their divisor count
for (int i = 0; i < N; i++) {
A[i] = divisorcount(A[i]);
}
// Convert every element of B[]
// to their divisor count
for (int i = 0; i < M; i++) {
B[i] = divisorcount(B[i]);
}
// Sort both the arrays
sort(A, A + N);
sort(B, B + M);
int winA = 0;
for (int i = 0; i < N; i++) {
int val = A[i];
int start = 0;
int end = M - 1;
int index = -1;
// For every element of A apply Binary Search
// to find number of pairs where A wins
while (start <= end) {
int mid = (start + end) / 2;
if (B[mid] <= val) {
index = mid;
start = mid + 1;
}
else {
end = mid - 1;
}
}
winA += (index + 1);
}
// B wins if A doesnot win
int winB = N * M - winA;
if (winA > winB) {
return "A";
}
else if (winB > winA) {
return "B";
}
return "Draw";
}
// Driver code
int main()
{
int A[] = { 4, 12, 24 };
int N = sizeof(A) / sizeof(A[0]);
int B[] = { 25, 28, 13, 45 };
int M = sizeof(B) / sizeof(B[0]);
cout << findwinner(A, B, N, M);
return 0;
} Java
// Java implementation of the
// above approach
import java.util.*;
class GFG{
// Function to return the count
// of divisors of elem
static int divisorcount(int elem)
{
int ans = 0;
for (int i = 1;
i <= Math.sqrt(elem); i++)
{
if (elem % i == 0)
{
if (i * i == elem)
ans++;
else
ans += 2;
}
}
return ans;
}
// Function to return the
// winner of the game
static String findwinner(int A[], int B[],
int N, int M)
{
// Convert every element of A[]
// to their divisor count
for (int i = 0; i < N; i++)
{
A[i] = divisorcount(A[i]);
}
// Convert every element of B[]
// to their divisor count
for (int i = 0; i < M; i++)
{
B[i] = divisorcount(B[i]);
}
// Sort both the arrays
Arrays.sort(A);
Arrays.sort(B);
int winA = 0;
for (int i = 0; i < N; i++)
{
int val = A[i];
int start = 0;
int end = M - 1;
int index = -1;
// For every element of A
// apply Binary Search to
// find number of pairs
// where A wins
while (start <= end)
{
int mid = (start +
end) / 2;
if (B[mid] <= val)
{
index = mid;
start = mid + 1;
}
else
{
end = mid - 1;
}
}
winA += (index + 1);
}
// B wins if A doesnot
// win
int winB = N * M - winA;
if (winA > winB)
{
return "A";
}
else if (winB > winA)
{
return "B";
}
return "Draw";
}
// Driver code
public static void main(String[] args)
{
int A[] = {4, 12, 24};
int N = A.length;
int B[] = {25, 28, 13, 45};
int M = B.length;
System.out.print(findwinner(A, B, N, M));
}
}
// This code is contributed by ChitranayalPython3
# Python3 implementation of the approach
from math import *
# Function to return the count
# of divisors of elem
def divisorcount(elem):
ans = 0
for i in range(1, int(sqrt(elem)) + 1):
if (elem % i == 0):
if (i * i == elem):
ans += 1
else:
ans += 2
return ans
# Function to return the winner of the game
def findwinner(A, B, N, M):
# Convert every element of A[]
# to their divisor count
for i in range(N):
A[i] = divisorcount(A[i])
# Convert every element of B[]
# to their divisor count
for i in range(M):
B[i] = divisorcount(B[i])
# Sort both the arrays
A.sort()
B.sort()
winA = 0
for i in range(N):
val = A[i]
start = 0
end = M - 1
index = -1
# For every element of A[] apply
# binary search to find number
# of pairs where A wins
while (start <= end):
mid = (start + end) // 2
if (B[mid] <= val):
index = mid
start = mid + 1
else:
end = mid - 1
winA += (index + 1)
# B wins if A doesnot win
winB = N * M - winA
if (winA > winB):
return "A"
elif (winB > winA):
return "B"
return "Draw"
# Driver code
if __name__ == '__main__':
A = [ 4, 12, 24 ]
N = len(A)
B = [ 25, 28, 13, 45 ]
M = len(B)
print(findwinner(A, B, N, M))
# This code is contributed by himanshu77输出：
A