给定整数数组arr []和查询数组q [] 。
对于第i个查询, index = q [i] [0]和value = q [i] [1] 。该任务针对每个查询,更新arr [index] = arr [index] +值,然后返回数组中所有偶数元素的和。
例子:
Input: arr[] = {1, 2, 3, 4}, q[] = {{0, 1}, {1, -3}, {0, -4}, {3, 2}}
Output: 8 6 2 4
At the beginning, the array is {1, 2, 3, 4}.
After adding 1 to arr[0], the array becomes {2, 2, 3, 4} and the sum of even values is 2 + 2 + 4 = 8.
Add -3 to arr[1], arr[] = {2, -1, 3, 4} and the sum of even values is 2 + 4 = 6.
Add -4 to arr[0], arr[] = {-2, -1, 3, 4} and the sum of even values is -2 + 4 = 2.
Adding 2 to arr[3], arr[] = {-2, -1, 3, 6} and the sum of even values is -2 + 6 = 4.
Input: arr[] = {1, 2, 2, 2}, q[] = {{0, 1}, {1, 1}}
Output: 8 6
天真的方法:对于每个查询,更新数组中的值并计算数组中所有偶数值的总和。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the sum of even elements
// after updating value at given index
int EvenSum(vector& A, int index, int value)
{
// Add given value to A[index]
A[index] = A[index] + value;
// To store the sum of even elements
int sum = 0;
for (int i = 0; i < A.size(); i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
return sum;
}
// Function to print the result for every query
void BalanceArray(vector& A, vector >& Q)
{
// Resultant vector that stores
// the result for every query
vector ANS;
int i, sum;
for (i = 0; i < Q.size(); i++) {
int index = Q[i][0];
int value = Q[i][1];
// Get sum of even elements after updating
// value at given index
sum = EvenSum(A, index, value);
// Store sum for each query
ANS.push_back(sum);
}
// Print the result for every query
for (i = 0; i < ANS.size(); i++)
cout << ANS[i] << " ";
}
// Driver code
int main()
{
vector A = { 1, 2, 3, 4 };
vector > Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the sum of even elements
// after updating value at given index
static int EvenSum(int [] A, int index, int value)
{
// Add given value to A[index]
A[index] = A[index] + value;
// To store the sum of even elements
int sum = 0;
for (int i = 0; i < A.length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
return sum;
}
// Function to print the result for every query
static void BalanceArray(int [] A, int [][] Q)
{
// Resultant vector that stores
// the result for every query
int [] ANS = new int[Q.length];
int i, sum;
for (i = 0; i < Q.length; i++)
{
int index = Q[i][0];
int value = Q[i][1];
// Get sum of even elements after updating
// value at given index
sum = EvenSum(A, index, value);
// Store sum for each query
ANS[i] = sum;
}
// Print the result for every query
for (i = 0; i < ANS.length; i++)
System.out.print(ANS[i] + " ");
}
// Driver code
public static void main(String []args)
{
int [] A = { 1, 2, 3, 4 };
int [][] Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the sum of even elements
// after updating value at given index
static int EvenSum(int [] A, int index, int value)
{
// Add given value to A[index]
A[index] = A[index] + value;
// To store the sum of even elements
int sum = 0;
for (int i = 0; i < A.Length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
return sum;
}
// Function to print the result for every query
static void BalanceArray(int [] A, int [, ] Q)
{
// Resultant vector that stores
// the result for every query
int [] ANS = new int[Q.GetLength(0)];
int i, sum;
for (i = 0; i < Q.GetLength(0); i++)
{
int index = Q[i, 0];
int value = Q[i, 1];
// Get sum of even elements after updating
// value at given index
sum = EvenSum(A, index, value);
// Store sum for each query
ANS[i] = sum;
}
// Print the result for every query
for (i = 0; i < ANS.Length; i++)
Console.Write(ANS[i] + " ");
}
// Driver code
public static void Main()
{
int [] A = new int [] { 1, 2, 3, 4 };
int [, ] Q = new int[ , ] { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of the approach
# Function to return the sum of even elements
# after updating value at given index
def EvenSum(A, index, value):
# Add given value to A[index]
A[index] = A[index] + value
# To store the sum of even elements
sum = 0
for i in A:
# If current element is even
if (i % 2 == 0):
sum = sum + i
return sum
# Function to prthe result for every query
def BalanceArray(A,Q):
# Resultant vector that stores
# the result for every query
ANS = []
i, sum = 0, 0
for i in range(len(Q)):
index = Q[i][0]
value = Q[i][1]
# Get sum of even elements after updating
# value at given index
sum = EvenSum(A, index, value)
# Store sum for each query
ANS.append(sum)
# Print the result for every query
for i in ANS:
print(i, end = " ")
# Driver code
A = [1, 2, 3, 4]
Q = [ [0, 1 ],
[1, -3],
[0, -4],
[3, 2 ] ]
BalanceArray(A, Q)
# This code is contributed by mohit kumar 29
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the result for every query
void BalanceArray(vector& A, vector >& Q)
{
vector ANS;
int i, sum = 0;
for (i = 0; i < A.size(); i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.size(); i++) {
int index = Q[i][0];
int value = Q[i][1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS.push_back(sum);
}
// Print the result for every query
for (i = 0; i < ANS.size(); i++)
cout << ANS[i] << " ";
}
// Driver code
int main()
{
vector A = { 1, 2, 3, 4 };
vector > Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the result for every query
static void BalanceArray(int [] A, int [][] Q)
{
int [] ANS = new int [A.length];
int i, sum = 0;
for (i = 0; i < A.length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.length; i++)
{
int index = Q[i][0];
int value = Q[i][1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS[i]= sum;
}
// Print the result for every query
for (i = 0; i < ANS.length; i++)
System.out.print(ANS[i] + " ");
}
// Driver code
public static void main(String [] args)
{
int [] A = { 1, 2, 3, 4 };
int [][] Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of the approach
# Function to print the result for
# every query
def BalanceArray(A, Q) :
ANS = []
sum = 0
for i in range(len(A)) :
# If current element is even
if (A[i] % 2 == 0) :
sum += A[i];
for i in range(len(Q)) :
index = Q[i][0];
value = Q[i][1];
# If element is even then
# remove it from sum
if (A[index] % 2 == 0) :
sum -= A[index];
A[index] += value;
# If the value becomes even
# after updating
if (A[index] % 2 == 0) :
sum += A[index];
# Store sum for each query
ANS.append(sum);
# Print the result for every query
for i in range(len(ANS)) :
print(ANS[i], end = " ");
# Driver code
if __name__ == "__main__" :
A = [ 1, 2, 3, 4 ];
Q = [ [ 0, 1 ], [ 1, -3 ],
[ 0, -4 ], [ 3, 2 ]];
BalanceArray(A, Q);
# This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the result for every query
static void BalanceArray(int [] A, int [, ] Q)
{
int [] ANS = new int [A.Length];
int i, sum = 0;
for (i = 0; i < A.Length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.GetLength(0); i++)
{
int index = Q[i, 0];
int value = Q[i, 1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS[i]= sum;
}
// Print the result for every query
for (i = 0; i < ANS.Length; i++)
Console.Write(ANS[i] + " ");
}
// Driver code
public static void Main()
{
int [] A = { 1, 2, 3, 4 };
int [, ] Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
PHP
8 6 2 4
时间复杂度: O(n 2 )
高效方法:
- 计算arr []中的偶数值之和
- 现在,如果给定索引处的arr []的值是偶数,即arr [index]%2 = 0,则从总和中减去arr [index]。
- 将给定值添加到arr [index],即arr [index] = arr [index] +值。
- 现在再次检查arr [index]的值是否为偶数,如果是,则将arr [index]添加到总和中。
- 打印每个查询的总和值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the result for every query
void BalanceArray(vector& A, vector >& Q)
{
vector ANS;
int i, sum = 0;
for (i = 0; i < A.size(); i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.size(); i++) {
int index = Q[i][0];
int value = Q[i][1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS.push_back(sum);
}
// Print the result for every query
for (i = 0; i < ANS.size(); i++)
cout << ANS[i] << " ";
}
// Driver code
int main()
{
vector A = { 1, 2, 3, 4 };
vector > Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the result for every query
static void BalanceArray(int [] A, int [][] Q)
{
int [] ANS = new int [A.length];
int i, sum = 0;
for (i = 0; i < A.length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.length; i++)
{
int index = Q[i][0];
int value = Q[i][1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS[i]= sum;
}
// Print the result for every query
for (i = 0; i < ANS.length; i++)
System.out.print(ANS[i] + " ");
}
// Driver code
public static void main(String [] args)
{
int [] A = { 1, 2, 3, 4 };
int [][] Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of the approach
# Function to print the result for
# every query
def BalanceArray(A, Q) :
ANS = []
sum = 0
for i in range(len(A)) :
# If current element is even
if (A[i] % 2 == 0) :
sum += A[i];
for i in range(len(Q)) :
index = Q[i][0];
value = Q[i][1];
# If element is even then
# remove it from sum
if (A[index] % 2 == 0) :
sum -= A[index];
A[index] += value;
# If the value becomes even
# after updating
if (A[index] % 2 == 0) :
sum += A[index];
# Store sum for each query
ANS.append(sum);
# Print the result for every query
for i in range(len(ANS)) :
print(ANS[i], end = " ");
# Driver code
if __name__ == "__main__" :
A = [ 1, 2, 3, 4 ];
Q = [ [ 0, 1 ], [ 1, -3 ],
[ 0, -4 ], [ 3, 2 ]];
BalanceArray(A, Q);
# This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the result for every query
static void BalanceArray(int [] A, int [, ] Q)
{
int [] ANS = new int [A.Length];
int i, sum = 0;
for (i = 0; i < A.Length; i++)
// If current element is even
if (A[i] % 2 == 0)
sum = sum + A[i];
for (i = 0; i < Q.GetLength(0); i++)
{
int index = Q[i, 0];
int value = Q[i, 1];
// If element is even then
// remove it from sum
if (A[index] % 2 == 0)
sum = sum - A[index];
A[index] = A[index] + value;
// If the value becomes even after updating
if (A[index] % 2 == 0)
sum = sum + A[index];
// Store sum for each query
ANS[i]= sum;
}
// Print the result for every query
for (i = 0; i < ANS.Length; i++)
Console.Write(ANS[i] + " ");
}
// Driver code
public static void Main()
{
int [] A = { 1, 2, 3, 4 };
int [, ] Q = { { 0, 1 },
{ 1, -3 },
{ 0, -4 },
{ 3, 2 } };
BalanceArray(A, Q);
}
}
// This code is contributed by ihritik
的PHP
8 6 2 4
时间复杂度: O(n)
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