给定Q个查询，其中每个查询由两个数字L和R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。
Parity of a number refers to whether it contains an odd or even number of 1-bits. The number has Even Parity if it contains even number of 1-bits.
例子：
Input: Q = [ [1, 10], [121, 211] ]
Output:
33
7493
Explanation:
binary(1) = 01, parity = 1
binary(2) = 10, parity = 1
binary(3) = 11, parity = 2
binary(4) = 100, parity = 1
binary(5) = 101, parity = 2
binary(6) = 110, parity = 2
binary(7) = 111, parity = 3
binary(8) = 1000, parity = 1
binary(9) = 1001, parity = 2
binary(10) = 1010, parity = 2
From 1 to 10, 3, 5, 6, 9 and 10 are the Even Parity numbers. Therefore the sum is 33.
From 121 to 211 the sum of all the even parity numbers is 7493.
Input: Q = [ [ 10, 10 ], [ 258, 785 ], [45, 245], [ 1, 1000]]
Output:
10
137676
14595
250750
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。
- 初始化前缀数组pref [] 。
- 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[]。
- 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 coder001C#
// 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。