给定一个大小为N的数组arr[]和一个数组Q[][] ,其中每一行代表一个{ X, Y }形式的查询,每个查询的任务是找到索引X处存在的数组元素的总和, X + Y , X + 2 * Y + …
例子:
Input: arr[] = { 1, 2, 7, 5, 4 }, Q[][] = { { 2, 1 }, { 3, 2 } }
Output: 16 5
Explanation:
Query1: arr[2] + arr[2 + 1] + arr[2 + 2] = 7 + 5 + 4 = 16.
Query2: arr[3] = 5.
Input: arr[] = { 3, 6, 1, 8, 0 } Q[][] = { { 0, 2 } }
Output: 4
朴素的方法:解决这个问题的最简单的方法是遍历每个查询的数组并打印arr[x] + arr[x + y] + arr[x + 2 * y] + …
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
void querySum(int arr[], int N,
int Q[][2], int M)
{
// Iterate over each query
for (int i = 0; i < M; i++) {
int x = Q[i][0];
int y = Q[i][1];
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array and calculate
// the sum of the expression
while (x < N) {
// Update sum
sum += arr[x];
// Update x
x += y;
}
cout << sum << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][2] = { { 2, 1 }, { 3, 2 } };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(Q) / sizeof(Q[0]);
querySum(arr, N, Q, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int arr[], int N,
int Q[][], int M)
{
// Iterate over each query
for(int i = 0; i < M; i++)
{
int x = Q[i][0];
int y = Q[i][1];
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array and calculate
// the sum of the expression
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
System.out.print(sum + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][] = { { 2, 1 }, { 3, 2 } };
int N = arr.length;
int M = Q.length;
querySum(arr, N, Q, M);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach
# Function to Find the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
# for all queries
def querySum(arr, N, Q, M):
# Iterate over each query
for i in range(M):
x = Q[i][0]
y = Q[i][1]
# Stores the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
sum = 0
# Traverse the array and calculate
# the sum of the expression
while (x < N):
# Update sum
sum += arr[x]
# Update x
x += y
print(sum, end=" ")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 7, 5, 4 ];
Q = [ [ 2, 1 ], [3, 2 ] ]
N = len(arr)
M = len(Q)
querySum(arr, N, Q, M)
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int []arr, int N,
int [,]Q, int M)
{
// Iterate over each query
for(int i = 0; i < M; i++)
{
int x = Q[i, 0];
int y = Q[i, 1];
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array and calculate
// the sum of the expression
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
Console.Write(sum + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 7, 5, 4 };
int [,]Q = { { 2, 1 }, { 3, 2 } };
int N = arr.Length;
int M = Q.GetLength(0);
querySum(arr, N, Q, M);
}
}
// This code is contributed by shikhasingrajput
Javascript
C++
// C++ program for the above approach
#include
using namespace std;
const int sz = 20;
const int sqr = int(sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to sqrt(N).
void precomputeExpressionForAllVal(int arr[], int N,
int dp[sz][sqr])
{
// Iterate over all possible values of X
for (int i = N - 1; i >= 0; i--) {
// Precompute for all possible values
// of an expression such that y <= sqrt(N)
for (int j = 1; j <= sqrt(N); j++) {
// If i + j less than N
if (i + j < N) {
// Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j];
}
else {
// Update dp[i][j]
dp[i][j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
int querySum(int arr[], int N,
int Q[][2], int M)
{
// dp[x][y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int dp[sz][sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[][]
for (int i = 0; i < M; i++) {
int x = Q[i][0];
int y = Q[i][1];
// If y is less than or equal
// to sqrt(N)
if (y <= sqrt(N)) {
cout << dp[x][y] << " ";
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, arr[]
while (x < N) {
// Update sum
sum += arr[x];
// Update x
x += y;
}
cout << sum << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][2] = { { 2, 1 }, { 3, 2 } };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(Q) / sizeof(Q[0]);
querySum(arr, N, Q, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int sz = 20;
static int sqr = (int)(Math.sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to Math.sqrt(N).
static void precomputeExpressionForAllVal(int arr[],
int N,
int dp[][])
{
// Iterate over all possible values of X
for(int i = N - 1; i >= 0; i--)
{
// Precompute for all possible values
// of an expression such that y <= Math.sqrt(N)
for(int j = 1; j <= Math.sqrt(N); j++)
{
// If i + j less than N
if (i + j < N)
{
// Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j];
}
else
{
// Update dp[i][j]
dp[i][j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int arr[], int N,
int Q[][], int M)
{
// dp[x][y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int [][]dp = new int[sz][sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[][]
for(int i = 0; i < M; i++)
{
int x = Q[i][0];
int y = Q[i][1];
// If y is less than or equal
// to Math.sqrt(N)
if (y <= Math.sqrt(N))
{
System.out.print(dp[x][y] + " ");
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, arr[]
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
System.out.print(sum + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][] = { { 2, 1 }, { 3, 2 } };
int N = arr.length;
int M = Q.length;
querySum(arr, N, Q, M);
}
}
// This code is contributed by shikhasingrajput
Python3
# python program for the above approach
import math
sz = 20
sqr = int(math.sqrt(sz)) + 1
# Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
# for all possible values of X and Y, where Y is
# less than or equal to sqrt(N).
def precomputeExpressionForAllVal(arr, N, dp):
# Iterate over all possible values of X
for i in range(N - 1, -1, -1) :
# Precompute for all possible values
# of an expression such that y <= sqrt(N)
for j in range (1,int(math.sqrt(N)) + 1):
# If i + j less than N
if (i + j < N):
# Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j]
else:
# Update dp[i][j]
dp[i][j] = arr[i]
# Function to Find the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
# for all queries
def querySum(arr, N, Q, M):
# dp[x][y]: Stores sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
dp = [ [0 for x in range(sz)]for x in range(sqr)]
precomputeExpressionForAllVal(arr, N, dp)
# Traverse the query array, Q[][]
for i in range (0,M):
x = Q[i][0]
y = Q[i][1]
# If y is less than or equal
# to sqrt(N)
if (y <= math.sqrt(N)):
print(dp[x][y])
continue
# Stores the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
sum = 0
# Traverse the array, arr[]
while (x < N):
# Update sum
sum += arr[x]
# Update x
x += y
print(sum)
# Driver Code
arr = [ 1, 2, 7, 5, 4 ]
Q = [ [ 2, 1 ], [ 3, 2]]
N = len(arr)
M = len(Q[0])
querySum(arr, N, Q, M)
# This code is contributed by amreshkumar3.
C#
// C# program for the above approach
using System;
class GFG{
static int sz = 20;
static int sqr = (int)(Math.Sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to Math.Sqrt(N).
static void precomputeExpressionForAllVal(int []arr,
int N,
int [,]dp)
{
// Iterate over all possible values of X
for(int i = N - 1; i >= 0; i--)
{
// Precompute for all possible values
// of an expression such that y <= Math.Sqrt(N)
for(int j = 1; j <= Math.Sqrt(N); j++)
{
// If i + j less than N
if (i + j < N)
{
// Update dp[i,j]
dp[i, j] = arr[i] + dp[i + j, j];
}
else
{
// Update dp[i,j]
dp[i, j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int []arr, int N,
int [,]Q, int M)
{
// dp[x,y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int [,]dp = new int[sz, sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[,]
for(int i = 0; i < M; i++)
{
int x = Q[i, 0];
int y = Q[i, 1];
// If y is less than or equal
// to Math.Sqrt(N)
if (y <= Math.Sqrt(N))
{
Console.Write(dp[x, y] + " ");
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, []arr
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
Console.Write(sum + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 7, 5, 4 };
int [,]Q = { { 2, 1 }, { 3, 2 } };
int N = arr.Length;
int M = Q.GetLength(0);
querySum(arr, N, Q, M);
}
}
// This code is contributed by shikhasingrajput
Javascript
16 5
时间复杂度: O(|Q| * O(N))
辅助空间: O(1)
有效的方法:可以通过使用动态规划技术和平方根分解技术预先计算所有可能的{ X, Y }值的给定表达式的值来解决问题。以下是递推关系:
if i + j < N
dp[i][j] = dp[i + j][j] + arr[i]
Otherwise,
dp[i][j] = arr[i]
dp[i][j]: Stores the sum of the given expression where X = i, Y = j
请按照以下步骤解决问题:
- 初始化一个二维数组,比如dp[][] ,以存储X和Y 的所有可能值的表达式总和,其中Y小于或等于sqrt(N) 。
- 使用制表方法填充dp[][]数组。
- 遍历数组Q[][] 。对于每个查询,检查Q[i][1] 的值是否小于或等于sqrt(N) 。如果发现为真,则打印dp[Q[i][0]][Q[i][1]] 的值。
- 否则,使用上述简单方法计算表达式的值并打印计算值。
下面是我们方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
const int sz = 20;
const int sqr = int(sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to sqrt(N).
void precomputeExpressionForAllVal(int arr[], int N,
int dp[sz][sqr])
{
// Iterate over all possible values of X
for (int i = N - 1; i >= 0; i--) {
// Precompute for all possible values
// of an expression such that y <= sqrt(N)
for (int j = 1; j <= sqrt(N); j++) {
// If i + j less than N
if (i + j < N) {
// Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j];
}
else {
// Update dp[i][j]
dp[i][j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
int querySum(int arr[], int N,
int Q[][2], int M)
{
// dp[x][y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int dp[sz][sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[][]
for (int i = 0; i < M; i++) {
int x = Q[i][0];
int y = Q[i][1];
// If y is less than or equal
// to sqrt(N)
if (y <= sqrt(N)) {
cout << dp[x][y] << " ";
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, arr[]
while (x < N) {
// Update sum
sum += arr[x];
// Update x
x += y;
}
cout << sum << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][2] = { { 2, 1 }, { 3, 2 } };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(Q) / sizeof(Q[0]);
querySum(arr, N, Q, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int sz = 20;
static int sqr = (int)(Math.sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to Math.sqrt(N).
static void precomputeExpressionForAllVal(int arr[],
int N,
int dp[][])
{
// Iterate over all possible values of X
for(int i = N - 1; i >= 0; i--)
{
// Precompute for all possible values
// of an expression such that y <= Math.sqrt(N)
for(int j = 1; j <= Math.sqrt(N); j++)
{
// If i + j less than N
if (i + j < N)
{
// Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j];
}
else
{
// Update dp[i][j]
dp[i][j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int arr[], int N,
int Q[][], int M)
{
// dp[x][y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int [][]dp = new int[sz][sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[][]
for(int i = 0; i < M; i++)
{
int x = Q[i][0];
int y = Q[i][1];
// If y is less than or equal
// to Math.sqrt(N)
if (y <= Math.sqrt(N))
{
System.out.print(dp[x][y] + " ");
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, arr[]
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
System.out.print(sum + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 7, 5, 4 };
int Q[][] = { { 2, 1 }, { 3, 2 } };
int N = arr.length;
int M = Q.length;
querySum(arr, N, Q, M);
}
}
// This code is contributed by shikhasingrajput
蟒蛇3
# python program for the above approach
import math
sz = 20
sqr = int(math.sqrt(sz)) + 1
# Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
# for all possible values of X and Y, where Y is
# less than or equal to sqrt(N).
def precomputeExpressionForAllVal(arr, N, dp):
# Iterate over all possible values of X
for i in range(N - 1, -1, -1) :
# Precompute for all possible values
# of an expression such that y <= sqrt(N)
for j in range (1,int(math.sqrt(N)) + 1):
# If i + j less than N
if (i + j < N):
# Update dp[i][j]
dp[i][j] = arr[i] + dp[i + j][j]
else:
# Update dp[i][j]
dp[i][j] = arr[i]
# Function to Find the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
# for all queries
def querySum(arr, N, Q, M):
# dp[x][y]: Stores sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
dp = [ [0 for x in range(sz)]for x in range(sqr)]
precomputeExpressionForAllVal(arr, N, dp)
# Traverse the query array, Q[][]
for i in range (0,M):
x = Q[i][0]
y = Q[i][1]
# If y is less than or equal
# to sqrt(N)
if (y <= math.sqrt(N)):
print(dp[x][y])
continue
# Stores the sum of
# arr[x]+arr[x+y]+arr[x+2*y] + ...
sum = 0
# Traverse the array, arr[]
while (x < N):
# Update sum
sum += arr[x]
# Update x
x += y
print(sum)
# Driver Code
arr = [ 1, 2, 7, 5, 4 ]
Q = [ [ 2, 1 ], [ 3, 2]]
N = len(arr)
M = len(Q[0])
querySum(arr, N, Q, M)
# This code is contributed by amreshkumar3.
C#
// C# program for the above approach
using System;
class GFG{
static int sz = 20;
static int sqr = (int)(Math.Sqrt(sz)) + 1;
// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all possible values of X and Y, where Y is
// less than or equal to Math.Sqrt(N).
static void precomputeExpressionForAllVal(int []arr,
int N,
int [,]dp)
{
// Iterate over all possible values of X
for(int i = N - 1; i >= 0; i--)
{
// Precompute for all possible values
// of an expression such that y <= Math.Sqrt(N)
for(int j = 1; j <= Math.Sqrt(N); j++)
{
// If i + j less than N
if (i + j < N)
{
// Update dp[i,j]
dp[i, j] = arr[i] + dp[i + j, j];
}
else
{
// Update dp[i,j]
dp[i, j] = arr[i];
}
}
}
}
// Function to Find the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
// for all queries
static void querySum(int []arr, int N,
int [,]Q, int M)
{
// dp[x,y]: Stores sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int [,]dp = new int[sz, sqr];
precomputeExpressionForAllVal(arr, N, dp);
// Traverse the query array, Q[,]
for(int i = 0; i < M; i++)
{
int x = Q[i, 0];
int y = Q[i, 1];
// If y is less than or equal
// to Math.Sqrt(N)
if (y <= Math.Sqrt(N))
{
Console.Write(dp[x, y] + " ");
continue;
}
// Stores the sum of
// arr[x]+arr[x+y]+arr[x+2*y] + ...
int sum = 0;
// Traverse the array, []arr
while (x < N)
{
// Update sum
sum += arr[x];
// Update x
x += y;
}
Console.Write(sum + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 7, 5, 4 };
int [,]Q = { { 2, 1 }, { 3, 2 } };
int N = arr.Length;
int M = Q.GetLength(0);
querySum(arr, N, Q, M);
}
}
// This code is contributed by shikhasingrajput
Javascript
16 5
时间复杂度: O(N * sqrt(N) + |Q| * sqrt(N))
辅助空间: O(N * sqrt(N))
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