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📜  范围查询,用于查找所有偶数奇偶校验和的和

📅  最后修改于: 2021-05-25 02:09:15             🧑  作者: Mango

给定Q个查询,其中每个查询由两个数字LR组成,这两个数字表示范围[L,R] 。任务是找到位于给定范围[L,R]中的所有偶数奇偶校验码的总和

例子:

方法:
这个想法是使用一个前缀和数组。直到特定索引被预先计算并存储在数组pref []中之前,所有偶数奇偶校验码的总和才能使每个查询都能在O(1)时间内得到回答。

  1. 初始化前缀数组pref []
  2. 从1到N进行迭代,并检查数字是否具有奇偶校验:
    • 如果该数字是偶数奇偶校验数,则pref []的当前索引将存储到目前为止找到的偶数奇偶校验数的总和。
    • 别的PREF的当前索引[]是相同的PREF先前索引处的值[]。
  3. 对于Q查询可以按以下方式计算范围[L,R]的所有偶校验数之和:
    sum = pref[R] - pref[L - 1]
    

下面是上述方法的实现

C++
// C++ program to find the sum
// of all Even Parity numbers
// in the given range
  
#include 
using namespace std;
  
// pref[] array to precompute
// the sum of all Even
// Parity Numbers
int pref[100001] = { 0 };
  
// Function that returns true
// if count of set bits in
// x is even
int isEvenParity(int num)
{
    // Parity will store the
    // count of set bits
    int parity = 0;
    int x = num;
    while (x != 0) {
        if (x & 1)
            parity++;
        x = x >> 1;
    }
  
    if (parity % 2 == 0)
        return num;
    else
        return 0;
}
  
// Function to precompute the
// sum of all even parity
// numbers upto 100000
void preCompute()
{
    for (int i = 1; i < 100001; i++) {
  
        // isEvenParity()
        // return the number i
        // if i has even parity
        // else return 0
        pref[i] = pref[i - 1]
                  + isEvenParity(i);
    }
}
  
// Function to print sum
// for each query
void printSum(int L, int R)
{
    cout << (pref[R] - pref[L - 1])
         << endl;
}
  
// Function to print sum of all
// even parity numbers between
// [L, R]
void printSum(int arr[2][2], int Q)
{
  
    // Function that pre computes
    // the sum of all even parity
    // numbers
    preCompute();
  
    // Iterate over all Queries
    // to print sum
    for (int i = 0; i < Q; i++) {
        printSum(arr[i][0],
                 arr[i][1]);
    }
}
// Driver code
int main()
{
    // Queries
    int N = 2;
    int Q[2][2] = { { 1, 10 },
                    { 121, 211 } };
  
    // Function that print
    // the sum of all even parity
    // numbers in Range [L, R]
    printSum(Q, N);
  
    return 0;
}


Java
// Java program to find the sum
// of all Even Parity numbers
// in the given range
import java.io.*; 
import java.util.*; 
  
class GFG { 
      
// pref[] array to precompute
// the sum of all Even
// Parity Numbers
static int[] pref = new int[100001];
  
// Function that returns true
// if count of set bits in
// x is even
static int isEvenParity(int num)
{
      
    // Parity will store the
    // count of set bits
    int parity = 0;
    int x = num;
      
    while (x != 0)
    {
        if ((x & 1) == 1)
            parity++;
              
        x = x >> 1;
    }
      
    if (parity % 2 == 0)
        return num;
    else
        return 0;
}
  
// Function to precompute the
// sum of all even parity
// numbers upto 100000
static void preCompute()
{
    for(int i = 1; i < 100001; i++)
    {
  
       // isEvenParity()
       // return the number i
       // if i has even parity
       // else return 0
       pref[i] = pref[i - 1] + isEvenParity(i);
    }
}
  
// Function to print sum
// for each query
static void printSum(int L, int R)
{
    System.out.println(pref[R] - pref[L - 1]);
}
  
// Function to print sum of all
// even parity numbers between
// [L, R]
static void printSum(int arr[][], int Q)
{
      
    // Function that pre computes
    // the sum of all even parity
    // numbers
    preCompute();
  
    // Iterate over all Queries
    // to print sum
    for(int i = 0; i < Q; i++) 
    {
       printSum(arr[i][0], arr[i][1]);
    }
}
      
// Driver code 
public static void main(String[] args) 
{ 
      
    // Queries
    int N = 2;
    int[][] Q = { { 1, 10 }, 
                  { 121, 211 } };
  
    // Function that print
    // the sum of all even parity
    // numbers in Range [L, R]
    printSum(Q, N);
} 
}
  
// This code is contributed by coder001


C#
// C# program to find the sum
// of all Even Parity numbers
// in the given range
using System;
  
class GFG { 
      
// pref[] array to precompute
// the sum of all Even
// Parity Numbers
static int[] pref = new int[100001];
  
// Function that returns true
// if count of set bits in
// x is even
static int isEvenParity(int num)
{
      
    // Parity will store the
    // count of set bits
    int parity = 0;
    int x = num;
      
    while (x != 0)
    {
        if ((x & 1) == 1)
            parity++;
              
        x = x >> 1;
    }
      
    if (parity % 2 == 0)
        return num;
    else
        return 0;
}
  
// Function to precompute the
// sum of all even parity
// numbers upto 100000
static void preCompute()
{
    for(int i = 1; i < 100001; i++)
    {
          
       // isEvenParity()
       // return the number i
       // if i has even parity
       // else return 0
       pref[i] = pref[i - 1] + isEvenParity(i);
    }
}
  
// Function to print sum
// for each query
static void printSum(int L, int R)
{
    Console.WriteLine(pref[R] - pref[L - 1]);
}
  
// Function to print sum of all
// even parity numbers between
// [L, R]
static void printSum(int[,] arr, int Q)
{
      
    // Function that pre computes
    // the sum of all even parity
    // numbers
    preCompute();
  
    // Iterate over all Queries
    // to print sum
    for(int i = 0; i < Q; i++) 
    {
       printSum(arr[i, 0], arr[i, 1]);
    }
}
      
// Driver code 
public static void Main() 
{ 
      
    // Queries
    int N = 2;
    int[,] Q = { { 1, 10 }, 
                 { 121, 211 } };
  
    // Function that print
    // the sum of all even parity
    // numbers in Range [L, R]
    printSum(Q, N);
} 
}
  
// This code is contributed by AbhiThakur


输出:
33
7493

时间复杂度: O(N) ,其中N是查询中的最大元素。

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