给定数组arr [] ，任务是计算给定数组中无序对(arr [i]，arr [j])的数量，以使(arr [i] * arr [j])％10 9 + 7等于1 。
例子：
Input: arr[] = {2, 236426, 280311812, 500000004}
Output: 2
Explanation: Two such pairs from the given array are:
- (2 * 500000004) % 1000000007 = 1
- (236426 * 280311812) % 1000000007 = 1
Input: arr[] = {4434, 923094278, 6565}
Output: 1
天真的方法：解决问题的最简单方法是遍历数组并从给定数组生成所有可能的对。对于每对，计算其乘积为10 9 + 7 。如果发现等于1 ，则增加此类对的计数。最后，打印获得的最终计数。
时间复杂度： O(N 2 )
辅助空间： O(N)
高效方法：要优化上述方法，请使用以下属性，如果(arr [i] * arr [j])％1000000007 = 1 ，则arr [j]是模10 9 + 7下的arr [i]的模逆。永远是独一无二的。请按照以下步骤解决问题：
- 初始化Map哈希，以将每个元素的频率存储在数组arr []中。
- 初始化变量pairCount ，以存储所需对的计数。
- 遍历数组并计算modularInverse是逆ARR [I]的下10 9 + 7并且由散列[modularInverse]增加pairCount和减1 pairCount的计数，如果modularInverse被发现是等于给Arr [i]中。
- 最后，打印出对数Count / 2作为必需的答案，因为上述方法对每一对都进行了两次计数。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
#define MOD 1000000007
// Iterative Function to calculate (x^y) % MOD
long long int modPower(long long int x,
long long int y)
{
// Initialize result
long long int res = 1;
// Update x if it exceeds MOD
x = x % MOD;
// If x is divisible by MOD
if (x == 0)
return 0;
while (y > 0) {
// If y is odd
if (y & 1)
// Multiply x with res
res = (res * x) % MOD;
// y must be even now
y = y / 2;
x = (x * x) % MOD;
}
return res;
}
// Function to count number of pairs
// whose product modulo 1000000007 is 1
int countPairs(long long int arr[], int N)
{
// Stores the count of
// desired pairs
int pairCount = 0;
// Stores the frequencies of
// each array element
map hash;
// Traverse the array and update
// frequencies in hash
for (int i = 0; i < N; i++) {
hash[arr[i]]++;
}
for (int i = 0; i < N; i++) {
// Calculate modular inverse of
// arr[i] under modulo 1000000007
long long int modularInverse
= modPower(arr[i], MOD - 2);
// Update desired count of pairs
pairCount += hash[modularInverse];
// If arr[i] and its modular inverse
// is equal under modulo MOD
if (arr[i] == modularInverse) {
// Updating count of desired pairs
pairCount--;
}
}
// Return the final count
return pairCount / 2;
}
int main()
{
long long int arr[]
= { 2, 236426, 280311812, 500000004 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << countPairs(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
static final int MOD = 1000000007;
// Iterative Function to calculate (x^y) % MOD
static long modPower(long x, int y)
{
// Initialize result
long res = 1;
// Update x if it exceeds MOD
x = x % MOD;
// If x is divisible by MOD
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd
if (y % 2 == 1)
// Multiply x with res
res = (res * x) % MOD;
// y must be even now
y = y / 2;
x = (x * x) % MOD;
}
return res;
}
// Function to count number of pairs
// whose product modulo 1000000007 is 1
static int countPairs(long arr[], int N)
{
// Stores the count of
// desired pairs
int pairCount = 0;
// Stores the frequencies of
// each array element
HashMap hash = new HashMap<>();
// Traverse the array and update
// frequencies in hash
for(int i = 0; i < N; i++)
{
if (hash.containsKey(arr[i]))
{
hash.put(arr[i], hash.get(arr[i]) + 1);
}
else
{
hash.put(arr[i], 1);
}
}
for(int i = 0; i < N; i++)
{
// Calculate modular inverse of
// arr[i] under modulo 1000000007
long modularInverse = modPower(arr[i],
MOD - 2);
// Update desired count of pairs
if (hash.containsKey(modularInverse))
pairCount += hash.get(modularInverse);
// If arr[i] and its modular inverse
// is equal under modulo MOD
if (arr[i] == modularInverse)
{
// Updating count of desired pairs
pairCount--;
}
}
// Return the final count
return pairCount / 2;
}
// Driver code
public static void main(String[] args)
{
long arr[] = { 2, 236426, 280311812, 500000004 };
int N = arr.length;
System.out.print(countPairs(arr, N));
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program to implement
# the above approach
from collections import defaultdict
MOD = 1000000007
# Iterative Function to
# calculate (x^y) % MOD
def modPower(x, y):
# Initialize result
res = 1
# Update x if it exceeds
# MOD
x = x % MOD
# If x is divisible by
# MOD
if (x == 0):
return 0
while (y > 0):
# If y is odd
if (y & 1):
# Multiply x with res
res = (res * x) % MOD
# y must be even now
y = y // 2
x = (x * x) % MOD
return res
# Function to count number
# of pairs whose product
# modulo 1000000007 is 1
def countPairs(arr, N):
# Stores the count of
# desired pairs
pairCount = 0
# Stores the frequencies of
# each array element
hash1 = defaultdict(int)
# Traverse the array and
# update frequencies in hash
for i in range(N):
hash1[arr[i]] += 1
for i in range(N):
# Calculate modular inverse
# of arr[i] under modulo
# 1000000007
modularInverse = modPower(arr[i],
MOD - 2)
# Update desired count of pairs
pairCount += hash1[modularInverse]
# If arr[i] and its modular
# inverse is equal under
# modulo MOD
if (arr[i] == modularInverse):
# Updating count of
# desired pairs
pairCount -= 1
# Return the final count
return pairCount // 2
# Driver code
if __name__ == "__main__":
arr = [2, 236426,
280311812,
500000004]
N = len(arr)
print(countPairs(arr, N))
# This code is contributed by ChitranayalC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
static int MOD = 1000000007;
// Iterative Function to calculate (x^y) % MOD
static long modPower(long x, int y)
{
// Initialize result
long res = 1;
// Update x if it exceeds MOD
x = x % MOD;
// If x is divisible by MOD
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd
if (y % 2 == 1)
// Multiply x with res
res = (res * x) % MOD;
// y must be even now
y = y / 2;
x = (x * x) % MOD;
}
return res;
}
// Function to count number of pairs
// whose product modulo 1000000007 is 1
static int countPairs(long []arr, int N)
{
// Stores the count of
// desired pairs
int pairCount = 0;
// Stores the frequencies of
// each array element
Dictionary hash = new Dictionary();
// Traverse the array and update
// frequencies in hash
for(int i = 0; i < N; i++)
{
if (hash.ContainsKey(arr[i]))
{
hash.Add(arr[i], hash[arr[i]] + 1);
}
else
{
hash.Add(arr[i], 1);
}
}
for(int i = 0; i < N; i++)
{
// Calculate modular inverse of
// arr[i] under modulo 1000000007
long modularInverse = modPower(arr[i],
MOD - 2);
// Update desired count of pairs
if (hash.ContainsKey(modularInverse))
pairCount += hash[modularInverse];
// If arr[i] and its modular inverse
// is equal under modulo MOD
if (arr[i] == modularInverse)
{
// Updating count of desired pairs
pairCount--;
}
}
// Return the final count
return pairCount / 2;
}
// Driver code
public static void Main()
{
long []arr = { 2, 236426, 280311812, 500000004 };
int N = arr.Length;
Console.WriteLine(countPairs(arr, N));
}
}
// This code is contributed by bgangwar59 输出：
2时间复杂度： O(NlogN)
辅助空间： O(N)