该序列首先由从1到n的所有奇数组成，然后由从2到n的其余偶数组成。假设n为1000，则序列变为1 3 5 7….999 2 4 6….1000
给定一个范围(L，R)，我们需要找到给定范围内该序列的数字之和。
注意：此处的范围为(L，R)L和R包含在范围内
例子：
Input : n = 10
Range 1 6
Output : 27
Explanation:
Sequence is 1 3 5 7 9 2 4 6 8 10
Sum in range (2, 6)
= 1 + 3 + 5 + 7 + 9 + 2
= 27
Input : n = 5
Range 1 2
Output : 4
Explanation:
sequence is 1 3 5 2 4
sum = 1 + 3 = 4这个想法是首先找到左(不包括左)之前的数字总和，然后找到右(包括右)之前的数字总和。我们得到的结果为第二总和减去第一总和。
如何求和到极限?
我们首先计算有多少个奇数，然后使用公式对奇数自然数和和偶数自然数进行求和。
如何找到奇数计数?
- 如果n为奇数，则奇数为((n / 2)+ 1)
- 如果n为偶数，则奇数为(n / 2)
通过简单的观察，我们得到奇数个数为ceil(n / 2)。因此，偶数为n – ceil(n / 2)。
- 前N个奇数之和为(N ^ 2)
- 前N个偶数之和为(N ^ 2)+ N
对于给定的数字x，我们如何找到从1到x的序列之和?
假设x小于奇数个数。
- 然后我们简单地返回(x * x)
如果x更大，则为奇数
- var = x-odd;
- 这意味着我们需要第一个var偶数
- 我们返回(odd * odd)+(var * var)+ var;
C++
// CPP program to find sum in the given range in
// the sequence 1 3 5 7.....N 2 4 6...N-1
#include
using namespace std;
// For our convenience
#define ll long long
// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
ll sumTillX(ll x, ll n)
{
// number of odd numbers
ll odd = ceil(n / 2.0);
if (x <= odd)
return x * x;
// number of extra even
// numbers required
ll even = x - odd;
return ((odd * odd) + (even * even) + even);
}
int rangeSum(int N, int L, int R)
{
return sumTillX(R, N) - sumTillX(L-1, N);
}
// Driver code
int main()
{
ll N = 10, L = 1, R = 6;
cout << rangeSum(N, L, R);
return 0;
} Java
// Java program to find
// sum in the given
// range in the sequence
// 1 3 5 7.....N
// 2 4 6...N-1
class GFG {
// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
static double sumTillX(double x,
double n)
{
// number of odd numbers
double odd = Math.ceil(n / 2.0);
if (x <= odd)
return x * x;
// number of extra even
// numbers required
double even = x - odd;
return ((odd * odd) + (even *
even) + even);
}
static double rangeSum(double N,
double L,
double R)
{
return sumTillX(R, N) -
sumTillX(L-1, N);
}
// Driver Code
public static void main(String args[])
{
long N = 10, L = 1, R = 6;
int n = 101;
System.out.println((int)rangeSum(N, L, R));
}
}
// This code is contributed by Sam007Python 3
# Python 3 program to find sum in the
# given range in the sequence 1 3 5 7
# .....N 2 4 6...N-1
import math
# For our convenience
#define ll long long
# Function that returns sum in the
# range 1 to x in the sequence
# 1 3 5 7.....N 2 4 6...N-1
def sumTillX(x, n):
# number of odd numbers
odd = math.ceil(n / 2.0)
if (x <= odd):
return x * x;
# number of extra even
# numbers required
even = x - odd;
return ((odd * odd) +
(even * even) + even);
def rangeSum(N, L, R):
return (sumTillX(R, N) -
sumTillX(L-1, N));
# Driver code
N = 10
L = 1
R = 6
print(rangeSum(N, L, R))
# This code is contributed by
# SmithaC#
// C# program to find sum in the given
// range in the sequence 1 3 5 7.....N
// 2 4 6...N-1
using System;
public class GFG {
// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
static double sumTillX(double x, double n)
{
// number of odd numbers
double odd = Math.Ceiling(n / 2.0);
if (x <= odd)
return x * x;
// number of extra even
// numbers required
double even = x - odd;
return ((odd * odd) + (even * even)
+ even);
}
static double rangeSum(double N, double L,
double R)
{
return sumTillX(R, N) - sumTillX(L-1, N);
}
// Driver code
public static void Main()
{
long N = 10, L = 1, R = 6;
Console.Write(rangeSum(N, L, R));
}
}
// This code is contributed by Sam007.PHP
Javascript
输出：
27