给定一个正整数N，任务是找到数m，从而当数N除以M，商等于它的余数，即(⌊N/M⌋= N mod M)表示，其中所有的计数⌊ ⌋表示给定数字的底值。
例子：
Input: N = 8
Output: 2
Explanation: When 8 is divided by 3 and 7, it returns the same Quotient and Remainder.
8 / 3 = 8 % 3, 8 / 7 = 8 % 7. Therefore, the required answer is 2.
Input: N = 1000000000000
Output: 84
原始的方法：最简单的方法是基于这样的事实的M不能多于N为任何大于N，商将是零。而其与N的模将始终为N。因此，迭代从1到N的所有数字，并对满足给定条件的所有此类数字进行计数。
时间复杂度： O(N)
辅助空间： O(1)
高效方法：最佳想法是基于以下陈述的观察结果：
If (⌊N/M⌋ = N mod M), then M + 1 must be a divisor of N.
观察证明：
If ⌊N/M⌋ = N mod M, then let N / M be equal to K.
Therefore, N must be equal to K * M + K as K is the quotient as well as the remainder.
N = K * M + K
N = K * (M + 1)
Therefore, for M to be in our answer set, it is necessary that M + 1 is a divisor of N.
Note that M + 1 must be a divisor of N is a necessary condition but not a sufficient condition for ⌊N/M⌋ = N mod M.
请按照以下步骤解决问题：
- 查找N的所有除数，并将它们存储在数组中。这可以以O(√N)的复杂度来计算。
- 通过遍历数组检查所有除数，如果除数减1满足给定条件⌊N/M⌋= N mod M ，则增加计数。
下面是上述方法的实现：
C++
// C++ program of the above approach
#include
using namespace std;
// Function to calculate the
// count of numbers such that it
// gives same quotient and remainder
void countDivisors(long long int n)
{
int count = 0;
// Stores divisor of number N.
vector divisor;
// Iterate through numbers from 2
// to sqrt(N) and store divisors of N
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i)
divisor.push_back(i);
else {
divisor.push_back(i);
divisor.push_back(n / i);
}
}
}
// As N is also divisor of itself
divisor.push_back(n);
// Iterate through divisors
for (auto x : divisor) {
x -= 1;
// Checking whether x satisfies the
// required condition
if ((n / x) == (n % x))
count++;
}
// Print the count
cout << count;
}
// Driver Code
int main()
{
// Given N
long long int N = 1000000000000;
// Function Call
countDivisors(N);
return 0;
} Java
// Java program of the above approach
class GFG {
// Function to calculate the
// count of numbers such that it
// gives same quotient and remainder
static void countDivisors(int n)
{
int count = 0;
int j = 0;
// Stores divisor of number N.
int divisor[] = new int[n];
// Iterate through numbers from 2
// to sqrt(N) and store divisors of N
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i){
divisor[j] = i;
j += 1;
}
else {
divisor[j] = i;
divisor[j + 1] = n / i;
j += 2;
}
}
}
// As N is also divisor of itself
divisor[j] = n;
// Iterate through divisors
for (int i = 0; i <= j; i++)
{
int x = divisor[i];
x -= 1;
// Checking whether x satisfies the
// required condition
if ((n / x) == (n % x))
count++;
}
// Print the count
System.out.print(count);
}
// Driver Code
public static void main (String[] args)
{
// Given N
int N = 10000000;
// Function Call
countDivisors(N);
}
}
// This code is contributed by AnkThonPython3
# Python3 program of the above approach
from math import floor, sqrt
# Function to calculate the
# count of numbers such that it
# gives same quotient and remainder
def countDivisors(n):
count = 0
# Stores divisor of number N.
divisor = []
# Iterate through numbers from 2
# to sqrt(N) and store divisors of N
for i in range(2, floor(sqrt(n))):
if (n % i == 0):
if (n // i == i):
divisor.append(i)
else:
divisor.append(i)
divisor.append(n // i)
# As N is also divisor of itself
divisor.append(n)
# Iterate through divisors
for x in divisor:
x -= 1
# Checking whether x satisfies the
# required condition
if ((n // x) == (n % x)):
count += 1
# Print the count
print(count)
# Driver Code
if __name__ == '__main__':
# Given N
N = 1000000000000
# Function Call
countDivisors(N)
# This code is contributed by mohit kumar 29C#
// C# program of the above approach
using System;
class GFG {
// Function to calculate the
// count of numbers such that it
// gives same quotient and remainder
static void countDivisors(int n)
{
int count = 0;
int j = 0;
// Stores divisor of number N.
int []divisor = new int[n];
// Iterate through numbers from 2
// to sqrt(N) and store divisors of N
for (int i = 2; i <= Math.Sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i){
divisor[j] = i;
j += 1;
}
else {
divisor[j] = i;
divisor[j + 1] = n / i;
j += 2;
}
}
}
// As N is also divisor of itself
divisor[j] = n;
// Iterate through divisors
for (int i = 0; i <= j; i++)
{
int x = divisor[i];
x -= 1;
// Checking whether x satisfies the
// required condition
if ((n / x) == (n % x))
count++;
}
// Print the count
Console.Write(count);
}
// Driver Code
public static void Main(String[] args)
{
// Given N
int N = 10000000;
// Function Call
countDivisors(N);
}
}
// This code contributed by shikhasingrajput84时间复杂度： O(sqrt(N))
辅助空间： O(sqrt(N))