给定一个数组arr[] ，其中最初所有元素都是0 ，另一个数组Q[][]包含K 个查询，其中每个查询代表一个范围[L, R] ，任务是向每个子数组添加1 ，其中每个子数组被定义通过范围[L, R] ，并返回所有唯一元素的总和。
注意：在Q[][]数组中使用从一开始的索引来表示范围。
例子：
Input: arr[] = { 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}}
Output: 6
Explanation:
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0 }.
Query 1: arr[] = { 1, 1, 1, 0, 0, 0 }.
Query 2: arr[] = { 1, 1, 1, 1, 1, 1 }.
Query 3: arr[] = { 1, 1, 2, 2, 1, 1 }.
Query 4: arr[] = { 1, 1, 3, 2, 1, 1 }.
Hence unique elements are {1, 3, 2}. Thus sum = 1 + 3 + 2 = 6.
Input: arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 4}, {5, 5}, {7, 8}, {8, 8}}
Output: 3
Explanation:
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }.
After processing all queries, arr[] = { 1, 1, 1, 1, 1, 0, 1, 2 }.
Hence unique elements are {1, 0, 2}. Thus sum = 1 + 0 + 2 = 3.
方法：思想是在L和R + 1索引处分别将值递增1并将数组递减1以处理每个查询。然后，计算数组的前缀和以找到 Q 查询后的最终数组。如本文所述。最后，借助哈希映射计算唯一元素的总和。
下面是上述方法的实现：
C++
// C++ implementation to find the
// sum of all unique elements of
// the array after Q queries
#include
using namespace std;
// Function to find the sum of
// unique elements after Q Query
int uniqueSum(int A[], int R[][2],
int N, int M)
{
// Updating the array after
// processing each query
for (int i = 0; i < M; ++i) {
int l = R[i][0], r = R[i][1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the final array
for (int i = 1; i < N; ++i) {
A[i] += A[i - 1];
}
// Variable to store the sum
int ans = 0;
// Hash to maintain perviously
// occured elements
unordered_set s;
// Loop to find the maximum sum
for (int i = 0; i < N; ++i) {
if (s.find(A[i]) == s.end())
ans += A[i];
s.insert(A[i]);
}
return ans;
}
// Driver code
int main()
{
int A[] = { 0, 0, 0, 0, 0, 0 };
int R[][2]
= { { 1, 3 }, { 4, 6 },
{ 3, 4 }, { 3, 3 } };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(R) / sizeof(R[0]);
cout << uniqueSum(A, R, N, M);
return 0;
} Java
// Java implementation to find the
// sum of all unique elements of
// the array after Q queries
import java.util.*;
class GFG{
// Function to find the sum of
// unique elements after Q Query
static int uniqueSum(int A[], int R[][],
int N, int M)
{
// Updating the array after
// processing each query
for (int i = 0; i < M; ++i)
{
int l = R[i][0], r = R[i][1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the final array
for (int i = 1; i < N; ++i)
{
A[i] += A[i - 1];
}
// Variable to store the sum
int ans = 0;
// Hash to maintain perviously
// occured elements
HashSet s = new HashSet();
// Loop to find the maximum sum
for (int i = 0; i < N; ++i)
{
if (!s.contains(A[i]))
ans += A[i];
s.add(A[i]);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int A[] = { 0, 0, 0, 0, 0, 0 };
int R[][] = { { 1, 3 }, { 4, 6 },
{ 3, 4 }, { 3, 3 } };
int N = A.length;
int M = R.length;
System.out.print(uniqueSum(A, R, N, M));
}
}
// This code is contributed by gauravrajput1 Python 3
# Python implementation to find the
# sum of all unique elements of
# the array after Q queries
# Function to find the sum of
# unique elements after Q Query
def uniqueSum(A, R, N, M) :
# Updating the array after
# processing each query
for i in range(0, M) :
l = R[i][0]
r = R[i][1] + 1
# Making it to 0-indexing
l -= 1
r -= 1
A[l] += 1
if (r < N) :
A[r] -= 1
# Iterating over the array
# to get the final array
for i in range(1, N) :
A[i] += A[i - 1]
# Variable to store the sum
ans = 0
# Hash to maintain perviously
# occured elements
s = {chr}
# Loop to find the maximum sum
for i in range(0, N) :
if (A[i] not in s) :
ans += A[i]
s.add(A[i])
return ans
# Driver code
A = [ 0, 0, 0, 0, 0, 0 ]
R = [ [ 1, 3 ], [ 4, 6 ],
[ 3, 4 ], [ 3, 3 ] ]
N = len(A)
M = len(R)
print(uniqueSum(A, R, N, M))
# This code is contributed by Sanjit_PrasadC#
// C# implementation to find the
// sum of all unique elements of
// the array after Q queries
using System;
using System.Collections.Generic;
class GFG{
// Function to find the sum of
// unique elements after Q Query
static int uniqueSum(int []A, int [,]R,
int N, int M)
{
// Updating the array after
// processing each query
for(int i = 0; i < M; ++i)
{
int l = R[i, 0], r = R[i, 1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the readonly array
for(int i = 1; i < N; ++i)
{
A[i] += A[i - 1];
}
// Variable to store the sum
int ans = 0;
// Hash to maintain perviously
// occured elements
HashSet s = new HashSet();
// Loop to find the maximum sum
for(int i = 0; i < N; ++i)
{
if (!s.Contains(A[i]))
ans += A[i];
s.Add(A[i]);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []A = { 0, 0, 0, 0, 0, 0 };
int [,]R = { { 1, 3 }, { 4, 6 },
{ 3, 4 }, { 3, 3 } };
int N = A.Length;
int M = R.GetLength(0);
Console.Write(uniqueSum(A, R, N, M));
}
}
// This code is contributed by Princi Singh Javascript
输出：
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