给定一个数组arr [],其中包含N个元素,其中N个元素同时包含正负元素,则任务是查找连续子序列的总数,这些子序列的乘积可以表示为两个整数的平方之差。
例子:
Input: arr[] = {1, 0, 2, 4, 5}
Output: 14
Explanation:
There are 14 subsequences whose product can be expressed as the difference of the square of two integers.
They are: {1}, {0}, {4}, {5}, {1, 0}, {1, 0, 2}, {1, 0, 2, 4}, {1, 0, 2, 4, 5}, {0, 2}, {0, 2, 4}, {0, 2, 4, 5}, {2, 4}, {2, 4, 5}, {4, 5}
The product of all the subsequences can be expressed as the difference of two squares. For example:
1 -> 1^2 – 0^2
0 -> 1^2 – 1^2
4 -> 2^2 – 0^2
5 -> 3^2 – 2^2
8 -> 3^2 – 1^2 …… and so on.
Input: arr[] = {-2, -7, 8, 9}
Output: 8
Explanation:
There are 8 subsequences whose product can be expressed as the difference of the square of two integers.
They are: {-7}, {8}, {9}, {-2, -7, 8}, {-2, -7, 8, 9}, {-7, 8}, {-7, 8, 9}, {8, 9}
The product of all the subsequences can be expressed as the difference of two squares. For example:
-7 -> 3^2 – 4^2
8 -> 3^2 – 1^2
9 -> 3^2 – 0^2
112 -> 11^2 – 3^2 …… and so on.
天真的方法:这个问题的天真的方法是生成所有连续的子序列并计算其乘积,然后简单地检查该数字是否可以表示为两个平方的差。
下面是上述方法的实现:
C++
// C++ implementation to count the
// number of contiguous subsequences
// whose product can be expressed as
// the square of difference of two integers
#include
using namespace std;
// Function to count the number
// of contiguous subsequences
// whose product can be expressed
// as square of difference of two integers
int CntcontSubs(int a[], int n)
{
int c = 0, d = 0, i, sum = 1, j;
// Iterating through the array
for (i = 0; i < n; i++) {
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (a[i] % 2 != 0 || a[i] % 4 == 0)
d++;
// Variable to compute the product
sum = a[i];
// Finding the remaining subsequences
for (j = i + 1; j < n; j++) {
sum = sum * a[j];
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (sum % 2 != 0 || sum % 4 == 0)
c++;
}
sum = 1;
}
// Return the number of subsequences
return c + d;
}
// Driver code
int main()
{
int arr[] = { 5, 4, 2, 9, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << CntcontSubs(arr, n);
return 0;
}
Java
// Java implementation to count the
// number of contiguous subsequences
// whose product can be expressed as
// the square of difference of two integers
class GFG{
// Function to count the number
// of contiguous subsequences
// whose product can be expressed
// as square of difference of two integers
static int CntcontSubs(int a[], int n)
{
int c = 0, d = 0, i, sum = 1, j;
// Iterating through the array
for (i = 0; i < n; i++) {
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (a[i] % 2 != 0 || a[i] % 4 == 0)
d++;
// Variable to compute the product
sum = a[i];
// Finding the remaining subsequences
for (j = i + 1; j < n; j++) {
sum = sum * a[j];
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (sum % 2 != 0 || sum % 4 == 0)
c++;
}
sum = 1;
}
// Return the number of subsequences
return c + d;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 5, 4, 2, 9, 8 };
int n = arr.length;
System.out.print(CntcontSubs(arr, n));
}
}
// This code contributed by PrinciRaj1992
Python3
# Python3 implementation to count the
# number of contiguous subsequences
# whose product can be expressed as
# the square of difference of two integers
# Function to count the number
# of contiguous subsequences
# whose product can be expressed
# as square of difference of two integers
def CntcontSubs(a, n):
c = 0
d = 0
sum = 1
# Iterating through the array
for i in range(n):
# Check if that number can be
# expressed as the square of
# difference of two numbers
if (a[i] % 2 != 0 or a[i] % 4 == 0):
d += 1
# Variable to compute the product
sum = a[i]
# Finding the remaining subsequences
for j in range(i + 1, n):
sum = sum * a[j]
# Check if that number can be
# expressed as the square of
# difference of two numbers
if (sum % 2 != 0 or sum % 4 == 0):
c += 1
sum = 1
# Return the number of subsequences
return c + d
# Driver code
if __name__ == '__main__':
arr=[5, 4, 2, 9, 8]
n = len(arr)
print(CntcontSubs(arr, n))
# This code is contributed by mohit kumar 29
C#
// C# implementation to count the
// number of contiguous subsequences
// whose product can be expressed as
// the square of difference of two integers
using System;
class GFG{
// Function to count the number
// of contiguous subsequences
// whose product can be expressed
// as square of difference of two integers
static int CntcontSubs(int []a, int n)
{
int c = 0, d = 0, i, sum = 1, j;
// Iterating through the array
for(i = 0; i < n; i++)
{
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (a[i] % 2 != 0 || a[i] % 4 == 0)
d++;
// Variable to compute the product
sum = a[i];
// Finding the remaining subsequences
for(j = i + 1; j < n; j++)
{
sum = sum * a[j];
// Check if that number can be
// expressed as the square of
// difference of two numbers
if (sum % 2 != 0 || sum % 4 == 0)
c++;
}
sum = 1;
}
// Return the number of subsequences
return c + d;
}
// Driver code
static void Main()
{
int []arr = { 5, 4, 2, 9, 8 };
int n = arr.Length;
Console.Write(CntcontSubs(arr, n));
}
}
// This code is contributed by grand_master
Javascript
C++
// C++ implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
#include
using namespace std;
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
int CntcontSubs(int a[], int n)
{
int prod = 1;
// Creating vectors to store
// the remainders and the
// subsequences
vector > vect;
vect.push_back(make_pair(0, 2));
vector two, zero;
// Iterating through the array
for (int i = 0; i < n; i++) {
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.push_back(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.push_back(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.push_back(make_pair(i + 1, a[i]));
}
vect.push_back(make_pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.empty())
return total;
else {
int sum = 0;
int pos1 = -1, pos2 = -1, pos3 = -1;
int sz = vect.size();
// Iterating through the vector
for (int i = 1; i + 1 < sz; i++) {
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect[i].second == 2) {
sum += (vect[i].first
- vect[i - 1].first)
* (vect[i + 1].first
- vect[i].first)
- 1;
}
}
// Returning the count
return total - sum - two.size();
}
}
// Driver code
int main()
{
int a[] = { 5, 4, 2, 9, 8 };
int n = sizeof(a) / sizeof(a[0]);
cout << CntcontSubs(a, n);
return 0;
}
Java
// Java implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
static int CntcontSubs(int a[], int n)
{
int prod = 1;
// Creating vectors to store
// the remainders and the
// subsequences
Vector vect = new Vector();
vect.add(new pair(0, 2));
Vector two = new Vector();
Vector zero = new Vector();
// Iterating through the array
for (int i = 0; i < n; i++)
{
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.add(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.add(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.add(new pair(i + 1, a[i]));
}
vect.add(new pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.isEmpty())
return total;
else
{
int sum = 0;
int pos1 = -1, pos2 = -1, pos3 = -1;
int sz = vect.size();
// Iterating through the vector
for (int i = 1; i + 1 < sz; i++)
{
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect.get(i).second == 2)
{
sum += (vect.get(i).first -
vect.get(i-1).first) *
(vect.get(i+1).first -
vect.get(i).first) - 1;
}
}
// Returning the count
return total - sum - two.size();
}
}
// Driver code
public static void main(String[] args)
{
int a[] = {5, 4, 2, 9, 8};
int n = a.length;
System.out.print(CntcontSubs(a, n));
}
}
// This code is contributed by shikhasingrajput
Python3
# Python3 implementation to count all the
# contiguous subsequences whose product is
# expressed as the square of the difference
# of two integers
# Function to count all the
# contiguous subsequences whose
# product is expressed as the square
# of the difference of two integers
def CntcontSubs(a, n):
prod = 1
# Creating vectors to store
# the remainders and the
# subsequences
vect = []
vect.append((0, 2))
two, zero = [], []
# Iterating through the array
for i in range(n):
# Finding the remainder when the
# element is divided by 4
a[i] = a[i] % 4
# Bringing all the elements in
# the range [0, 3]
if (a[i] < 0):
a[i] = a[i] + 4
# If the remainder is 2, store
# the index of the
if (a[i] == 2):
two.append(i + 1)
# If the remainder is 2, store
# the index of the
if (a[i] == 0):
zero.append(i + 1)
if (a[i] == 0 or a[i] == 2):
vect.append((i + 1, a[i]))
vect.append((n + 1, 2))
# Finding the total number of subsequences
total = (n * (n + 1)) // 2
# If there are no numbers which
# yield the remainder 2
if (len(two) == 0):
return total
else:
Sum = 0
pos1, pos2, pos3 = -1, -1, -1
sz = len(vect)
# Iterating through the vector
for i in range(1, sz - 1):
# If the element is 2, find the
# nearest 2 or 0 and find the
# number of elements between them
if (vect[i][1] == 2) :
Sum += ((vect[i][0] - vect[i - 1][0]) *
(vect[i + 1][0] - vect[i][0]) - 1)
# Returning the count
return (total - Sum - len(two))
# Driver Code
a = [ 5, 4, 2, 9, 8 ]
n = len(a)
print(CntcontSubs(a, n))
# This code is contributed by divyeshrabadiya07
C#
// C# implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
static int CntcontSubs(int []a, int n)
{
// Creating vectors to store
// the remainders and the
// subsequences
List vect = new List();
vect.Add(new pair(0, 2));
List two = new List();
List zero = new List();
// Iterating through the array
for(int i = 0; i < n; i++)
{
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.Add(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.Add(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.Add(new pair(i + 1, a[i]));
}
vect.Add(new pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.Count == 0)
return total;
else
{
int sum = 0;
int sz = vect.Count;
// Iterating through the vector
for(int i = 1; i + 1 < sz; i++)
{
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect[i].second == 2)
{
sum += (vect[i].first -
vect[i - 1].first) *
(vect[i + 1].first -
vect[i].first) - 1;
}
}
// Returning the count
return total - sum - two.Count;
}
}
// Driver code
public static void Main(String[] args)
{
int []a = { 5, 4, 2, 9, 8 };
int n = a.Length;
Console.Write(CntcontSubs(a, n));
}
}
// This code is contributed by Amit Katiyar
13
时间复杂度: O(N 2 ) ,其中N是数组的长度。
有效的方法:这种想法背后的恒等式是,如果一个数字除以4则得到2的余数,则数字不能表示为两个平方的差。因此,该想法是找到所有产生2乘积并减去的子序列。这些来自数组的所有可能子序列。连续子序列的总数可以通过以下公式获得: (N *(N + 1))/ 2
- 如果数组包含元素0,则包含该元素的所有子序列的乘积变为0。因此,所有这些子序列都可以表示为两个平方的差。
- 如果在数字除以4时有任何元素将余数2赋给剩余数,则必须避免所有最近的2或0的子序列,因为当遇到2时余数变为4,而当遇到0时余数变为0。
例子:
- 令arr [] = {6、5、13、10、4、8、14、17}。
- 当元素除以4时,我们将计算余数。因此,{2,1,1,2,2,0,0,2,1}是给定数组的余数。
- 在这里,我们得到索引1、4、7的余数2。让我们观察第一个索引的2。
- 以下是余数数组{2},{2、1},{2、1、1},{2、1、1、1、2},{2、1、1、2、0}…{ 2、1、1、2、0、0、2、1}。
- 子序列的乘积为{2,2,2,2,4,0…。 0}。
- 因此,我们得到从索引1到索引3的乘积为2。因此,从索引1到索引3(乘积为2)的总连续子序列为2(即[index-3 – index-1])。
- 因此,很明显,我们找到元素2或0的最近索引,并忽略所有子序列,直到该索引为止。
下面是上述方法的实现:
C++
// C++ implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
#include
using namespace std;
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
int CntcontSubs(int a[], int n)
{
int prod = 1;
// Creating vectors to store
// the remainders and the
// subsequences
vector > vect;
vect.push_back(make_pair(0, 2));
vector two, zero;
// Iterating through the array
for (int i = 0; i < n; i++) {
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.push_back(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.push_back(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.push_back(make_pair(i + 1, a[i]));
}
vect.push_back(make_pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.empty())
return total;
else {
int sum = 0;
int pos1 = -1, pos2 = -1, pos3 = -1;
int sz = vect.size();
// Iterating through the vector
for (int i = 1; i + 1 < sz; i++) {
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect[i].second == 2) {
sum += (vect[i].first
- vect[i - 1].first)
* (vect[i + 1].first
- vect[i].first)
- 1;
}
}
// Returning the count
return total - sum - two.size();
}
}
// Driver code
int main()
{
int a[] = { 5, 4, 2, 9, 8 };
int n = sizeof(a) / sizeof(a[0]);
cout << CntcontSubs(a, n);
return 0;
}
Java
// Java implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
static int CntcontSubs(int a[], int n)
{
int prod = 1;
// Creating vectors to store
// the remainders and the
// subsequences
Vector vect = new Vector();
vect.add(new pair(0, 2));
Vector two = new Vector();
Vector zero = new Vector();
// Iterating through the array
for (int i = 0; i < n; i++)
{
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.add(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.add(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.add(new pair(i + 1, a[i]));
}
vect.add(new pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.isEmpty())
return total;
else
{
int sum = 0;
int pos1 = -1, pos2 = -1, pos3 = -1;
int sz = vect.size();
// Iterating through the vector
for (int i = 1; i + 1 < sz; i++)
{
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect.get(i).second == 2)
{
sum += (vect.get(i).first -
vect.get(i-1).first) *
(vect.get(i+1).first -
vect.get(i).first) - 1;
}
}
// Returning the count
return total - sum - two.size();
}
}
// Driver code
public static void main(String[] args)
{
int a[] = {5, 4, 2, 9, 8};
int n = a.length;
System.out.print(CntcontSubs(a, n));
}
}
// This code is contributed by shikhasingrajput
Python3
# Python3 implementation to count all the
# contiguous subsequences whose product is
# expressed as the square of the difference
# of two integers
# Function to count all the
# contiguous subsequences whose
# product is expressed as the square
# of the difference of two integers
def CntcontSubs(a, n):
prod = 1
# Creating vectors to store
# the remainders and the
# subsequences
vect = []
vect.append((0, 2))
two, zero = [], []
# Iterating through the array
for i in range(n):
# Finding the remainder when the
# element is divided by 4
a[i] = a[i] % 4
# Bringing all the elements in
# the range [0, 3]
if (a[i] < 0):
a[i] = a[i] + 4
# If the remainder is 2, store
# the index of the
if (a[i] == 2):
two.append(i + 1)
# If the remainder is 2, store
# the index of the
if (a[i] == 0):
zero.append(i + 1)
if (a[i] == 0 or a[i] == 2):
vect.append((i + 1, a[i]))
vect.append((n + 1, 2))
# Finding the total number of subsequences
total = (n * (n + 1)) // 2
# If there are no numbers which
# yield the remainder 2
if (len(two) == 0):
return total
else:
Sum = 0
pos1, pos2, pos3 = -1, -1, -1
sz = len(vect)
# Iterating through the vector
for i in range(1, sz - 1):
# If the element is 2, find the
# nearest 2 or 0 and find the
# number of elements between them
if (vect[i][1] == 2) :
Sum += ((vect[i][0] - vect[i - 1][0]) *
(vect[i + 1][0] - vect[i][0]) - 1)
# Returning the count
return (total - Sum - len(two))
# Driver Code
a = [ 5, 4, 2, 9, 8 ]
n = len(a)
print(CntcontSubs(a, n))
# This code is contributed by divyeshrabadiya07
C#
// C# implementation to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to count all the
// contiguous subsequences whose
// product is expressed as the square
// of the difference of two integers
static int CntcontSubs(int []a, int n)
{
// Creating vectors to store
// the remainders and the
// subsequences
List vect = new List();
vect.Add(new pair(0, 2));
List two = new List();
List zero = new List();
// Iterating through the array
for(int i = 0; i < n; i++)
{
// Finding the remainder when the
// element is divided by 4
a[i] = a[i] % 4;
// Bringing all the elements in
// the range [0, 3]
if (a[i] < 0)
a[i] = a[i] + 4;
// If the remainder is 2, store
// the index of the
if (a[i] == 2)
two.Add(i + 1);
// If the remainder is 2, store
// the index of the
if (a[i] == 0)
zero.Add(i + 1);
if (a[i] == 0 || a[i] == 2)
vect.Add(new pair(i + 1, a[i]));
}
vect.Add(new pair(n + 1, 2));
// Finding the total number of subsequences
int total = (n * (n + 1)) / 2;
// If there are no numbers which
// yield the remainder 2
if (two.Count == 0)
return total;
else
{
int sum = 0;
int sz = vect.Count;
// Iterating through the vector
for(int i = 1; i + 1 < sz; i++)
{
// If the element is 2, find the nearest
// 2 or 0 and find the number of
// elements between them
if (vect[i].second == 2)
{
sum += (vect[i].first -
vect[i - 1].first) *
(vect[i + 1].first -
vect[i].first) - 1;
}
}
// Returning the count
return total - sum - two.Count;
}
}
// Driver code
public static void Main(String[] args)
{
int []a = { 5, 4, 2, 9, 8 };
int n = a.Length;
Console.Write(CntcontSubs(a, n));
}
}
// This code is contributed by Amit Katiyar
13
时间复杂度: O(N) ,其中N是数组的长度。