给定一个整数N ，任务是计算所有公共差等于1且所有元素之和等于N的算术级数。
例子：
Input: N = 12
Output: 3
Explanation:
Following three APs satisfy the required conditions:
- {3, 4, 5}
- {−2, −1, 0, 1, 2, 3, 4, 5}
- {−11, −10, −9, …, 10, 11, 12]}
Input: N = 963761198400
Output: 1919
方法：可以使用AP的以下属性来解决给定的问题：
- 具有第一项为A且共同差为1的N个项的AP系列的总和由公式S =(N / 2)*(2 * A + N -1)给出。
- 求解上面的方程式：
S = N / 2 [2 *A + N − 1].
Rearranging this and multiplying both sides by 2
=> 2 * S = N * (N + 2 * A − 1)
Rearranging further
=> A = ((2 * N / i ) − i + 1) / 2.
Now, iterate over all the factors of 2 * N, and check for each factor i, whether (2 * N/i) − i + 1 is even or not.
请按照以下步骤解决问题：
- 初始化一个变量，例如count，以存储在给定条件下可能的AP系列的数量。
- 遍历2 * N的所有因子，对于每个第i因子，检查(2 * N / i)− i + 1是否为偶数。
- 如果发现为真，则增加计数。
- 打印计数– 1作为必需答案，因为仅包含N的序列需要丢弃。
下面是上述方法的实现：
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to count all possible
// AP series with common difference
// 1 and sum of elements equal to N
void countAPs(long long int N)
{
// Stores the count of AP series
long long int count = 0;
// Traverse through all factors of 2 * N
for (long long int i = 1;
i * i <= 2 * N; i++) {
long long int res = 2 * N;
if (res % i == 0) {
// Check for the given conditions
long long int op = res / i - i + 1;
if (op % 2 == 0) {
// Increment count
count++;
}
if (i * i != res
and (i - res / i + 1) % 2 == 0) {
count++;
}
}
}
// Print count - 1
cout << count - 1 << "\n";
}
// Driver Code
int main()
{
// Given value of N
long long int N = 963761198400;
// Function call to count
// required number of AP series
countAPs(N);
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to count all possible
// AP series with common difference
// 1 and sum of elements equal to N
static void countAPs(long N)
{
// Stores the count of AP series
long count = 0;
// Traverse through all factors of 2 * N
for (long i = 1; i * i <= 2 * N; i++) {
long res = 2 * N;
if (res % i == 0) {
// Check for the given conditions
long op = res / i - i + 1;
if (op % 2 == 0) {
// Increment count
count++;
}
if (i * i != res
&& (i - res / i + 1) % 2 == 0) {
count++;
}
}
}
// Print count - 1
System.out.println(count - 1);
}
// Driver code
public static void main(String[] args)
{
// Given value of N
long N = 963761198400L;
// Function call to count
// required number of AP series
countAPs(N);
}
}
// This code is contributed by Kingash.Python3
# Python3 program to implement
# the above approach
# Function to count all possible
# AP series with common difference
# 1 and sum of elements equal to N
def countAPs(N) :
# Stores the count of AP series
count = 0
# Traverse through all factors of 2 * N
i = 1
while(i * i <= 2 * N) :
res = 2 * N
if (res % i == 0) :
# Check for the given conditions
op = res / i - i + 1
if (op % 2 == 0) :
# Increment count
count += 1
if (i * i != res
and (i - res / i + 1) % 2 == 0) :
count += 1
i += 1
# Prcount - 1
print(count - 1)
# Driver Code
# Given value of N
N = 963761198400
# Function call to count
# required number of AP series
countAPs(N)
# This code is contributed by sanjoy_62.C#
// C# program for above approach
using System;
public class GFG
{
// Function to count all possible
// AP series with common difference
// 1 and sum of elements equal to N
static void countAPs(long N)
{
// Stores the count of AP series
long count = 0;
// Traverse through all factors of 2 * N
for (long i = 1; i * i <= 2 * N; i++) {
long res = 2 * N;
if (res % i == 0) {
// Check for the given conditions
long op = res / i - i + 1;
if (op % 2 == 0) {
// Increment count
count++;
}
if (i * i != res
&& (i - res / i + 1) % 2 == 0) {
count++;
}
}
}
// Print count - 1
Console.WriteLine(count - 1);
}
// Driver code
public static void Main(String[] args)
{
// Given value of N
long N = 963761198400L;
// Function call to count
// required number of AP series
countAPs(N);
}
}
// This code is contributed by sanjoy_62.输出：
1919时间复杂度： O(√N)
辅助空间： O(1)