📌  相关文章
📜  以1 / n的十进制值查找周期长度

📅  最后修改于: 2021-04-23 19:28:15             🧑  作者: Mango

给定正整数n，以十进制值1 / n查找周期。以十进制值表示的期间是不断重复的位数(小数点后的某处)。
例子 ： 

Input:  n = 3
Output: 1
The value of 1/3 is 0.333333...

Input: n = 7
Output: 6
The value of 1/7 is 0.142857142857142857.....

Input: n = 210
Output: 6
The value of 1/210 is 0.0047619047619048.....

让我们首先讨论一个简单的问题，即找到值为1 / n的单个数字。
如何找到1 / n值的单个数字?
让我们以一个例子来了解该过程。例如，对于n =7。可以通过执行10/7获得第一个数字。第二个数字可以通过30/7来获得(上一个除法运算中的余数为3)。第三位数字可以通过20/7获得(2是上一个除法的余数)。因此，想法是获取第一个数字，然后继续将(remainder * 10)/ n的值用作下一个数字，并保持更新余数为(remainder * 10)％10。此处讨论了完整的程序。
如何找到时期?
1 / n的周期等于上述过程中使用的余数的周期。从数字直接取自余数的事实可以很容易地证明这一点。
关于余数序列的一个有趣的事实是，该余数序列周期中的所有燕鸥都是截然不同的。这样做的原因很简单，如果剩余部分重复出现，那就是新时期的开始。
以下是上述想法的实现。

CPP
// C++ program to find length of period of 1/n
#include 
#include 
using namespace std;
 
// Function to find length of period in 1/n
int getPeriod(int n)
{
   // Create a map to store mapping from remainder
   // its position
   map mymap;
   map::iterator it;
 
   // Initialize remainder and position of remainder
   int rem = 1, i = 1;
 
   // Keep finding remainders till a repeating remainder
   // is found
   while (true)
   {
        // Find next remainder
        rem = (10*rem) % n;
 
        // If remainder exists in mymap, then the difference
        // between current and previous position is length of
        // period
        it = mymap.find(rem);
        if (it != mymap.end())
            return (i - it->second);
 
        // If doesn't exist, then add 'i' to mymap
        mymap[rem] = i;
        i++;
   }
 
   // This code should never be reached
   return INT_MAX;
}
 
// Driver program to test above function
int main()
{
    cout <<  getPeriod(3) << endl;
    cout <<  getPeriod(7) << endl;
    return 0;
}

C++
// C++ program to find length of period of 1/n without
// using map or hash
#include 
using namespace std;
 
// Function to find length of period in 1/n
int getPeriod(int n)
{
   // Find the (n+1)th remainder after decimal point
   // in value of 1/n
   int rem = 1; // Initialize remainder
   for (int i = 1; i <= n+1; i++)
         rem = (10*rem) % n;
 
   // Store (n+1)th remainder
   int d = rem;
 
   // Count the number of remainders before next
   // occurrence of (n+1)'th remainder 'd'
   int count = 0;
   do {
      rem = (10*rem) % n;
      count++;
   } while(rem != d);
 
   return count;
}
 
// Driver program to test above function
int main()
{
    cout <<  getPeriod(3) << endl;
    cout <<  getPeriod(7) << endl;
    return 0;
}

Java
// Java program to find length
// of period of 1/n without using
// map or hash
 
class GFG {
     
// Function to find length of period in 1/n
static int getPeriod(int n)
{
    // Find the (n+1)th remainder after
    // decimal point in value of 1/n
     
    int rem = 1; // Initialize remainder
    for (int i = 1; i <= n + 1; i++)
            rem = (10 * rem) % n;
     
    // Store (n+1)th remainder
    int d = rem;
     
    // Count the number of remainders before next
    // occurrence of (n+1)'th remainder 'd'
    int count = 0;
    do {
        rem = (10 * rem) % n;
        count++;
    } while(rem != d);
     
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    System.out.println(getPeriod(3) );
    System.out.println(getPeriod(7));
}
}
 
// This code is contributed by Smitha Dinesh Semwal

Python3
# Python3 program to find length of
# period of 1/n without using map or hash
 
# Function to find length
# of period in 1/n
def getPeriod( n) :
 
    # Find the (n+1)th remainder after
    # decimal point in value of 1/n
    rem = 1 # Initialize remainder
    for i in range(1, n + 2):
        rem = (10 * rem) % n
 
    # Store (n+1)th remainder
    d = rem
     
    # Count the number of remainders
    # before next occurrence of
    # (n+1)'th remainder 'd'
    count = 0
    rem = (10 * rem) % n
    count += 1
    while rem != d :
        rem = (10 * rem) % n
        count += 1
     
    return count
 
# Driver Code
if __name__ == "__main__":
 
    print (getPeriod(3))
    print (getPeriod(7))
 
# This code is contributed
# by ChitraNayal

C#
// C# program to find length of
// period of 1/n without using
// map or hash
using System;
 
class GFG {
     
// Function to find length of period in 1/n
static int getPeriod(int n)
{
    // Find the (n+1)th remainder after
    // decimal point in value of 1/n
     
    int rem = 1; // Initialize remainder
    for (int i = 1; i <= n + 1; i++)
            rem = (10 * rem) % n;
     
    // Store (n+1)th remainder
    int d = rem;
     
    // Count the number of remainders before next
    // occurrence of (n+1)'th remainder 'd'
    int count = 0;
    do {
        rem = (10 * rem) % n;
        count++;
    } while(rem != d);
     
    return count;
}
 
// Driver code
public static void Main()
{
    Console.Write(getPeriod(3) + "\n");
    Console.Write(getPeriod(7));
}
}
 
// This code is contributed by Smitha Dinesh Semwal

PHP

Javascript

输出： 

1
6

利用以下事实，我们可以避免使用map或hash。对于数字n，最多可以有n个不同的余数。同样，该期间可能不是从第一个余数开始的，因为一些初始余数可能是非重复的(不是任何时期的一部分)。因此，要确保从一个周期中选取了余数，请从第(n + 1)个余数开始并继续寻找下一个余数。第(n + 1)次余数与下一次出现之间的距离是周期的长度。

C++ 

// C++ program to find length of period of 1/n without
// using map or hash
#include 
using namespace std;
 
// Function to find length of period in 1/n
int getPeriod(int n)
{
   // Find the (n+1)th remainder after decimal point
   // in value of 1/n
   int rem = 1; // Initialize remainder
   for (int i = 1; i <= n+1; i++)
         rem = (10*rem) % n;
 
   // Store (n+1)th remainder
   int d = rem;
 
   // Count the number of remainders before next
   // occurrence of (n+1)'th remainder 'd'
   int count = 0;
   do {
      rem = (10*rem) % n;
      count++;
   } while(rem != d);
 
   return count;
}
 
// Driver program to test above function
int main()
{
    cout <<  getPeriod(3) << endl;
    cout <<  getPeriod(7) << endl;
    return 0;
}

Java

// Java program to find length
// of period of 1/n without using
// map or hash
 
class GFG {
     
// Function to find length of period in 1/n
static int getPeriod(int n)
{
    // Find the (n+1)th remainder after
    // decimal point in value of 1/n
     
    int rem = 1; // Initialize remainder
    for (int i = 1; i <= n + 1; i++)
            rem = (10 * rem) % n;
     
    // Store (n+1)th remainder
    int d = rem;
     
    // Count the number of remainders before next
    // occurrence of (n+1)'th remainder 'd'
    int count = 0;
    do {
        rem = (10 * rem) % n;
        count++;
    } while(rem != d);
     
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    System.out.println(getPeriod(3) );
    System.out.println(getPeriod(7));
}
}
 
// This code is contributed by Smitha Dinesh Semwal

Python3 

# Python3 program to find length of
# period of 1/n without using map or hash
 
# Function to find length
# of period in 1/n
def getPeriod( n) :
 
    # Find the (n+1)th remainder after
    # decimal point in value of 1/n
    rem = 1 # Initialize remainder
    for i in range(1, n + 2):
        rem = (10 * rem) % n
 
    # Store (n+1)th remainder
    d = rem
     
    # Count the number of remainders
    # before next occurrence of
    # (n+1)'th remainder 'd'
    count = 0
    rem = (10 * rem) % n
    count += 1
    while rem != d :
        rem = (10 * rem) % n
        count += 1
     
    return count
 
# Driver Code
if __name__ == "__main__":
 
    print (getPeriod(3))
    print (getPeriod(7))
 
# This code is contributed
# by ChitraNayal

C＃ 

// C# program to find length of
// period of 1/n without using
// map or hash
using System;
 
class GFG {
     
// Function to find length of period in 1/n
static int getPeriod(int n)
{
    // Find the (n+1)th remainder after
    // decimal point in value of 1/n
     
    int rem = 1; // Initialize remainder
    for (int i = 1; i <= n + 1; i++)
            rem = (10 * rem) % n;
     
    // Store (n+1)th remainder
    int d = rem;
     
    // Count the number of remainders before next
    // occurrence of (n+1)'th remainder 'd'
    int count = 0;
    do {
        rem = (10 * rem) % n;
        count++;
    } while(rem != d);
     
    return count;
}
 
// Driver code
public static void Main()
{
    Console.Write(getPeriod(3) + "\n");
    Console.Write(getPeriod(7));
}
}
 
// This code is contributed by Smitha Dinesh Semwal

的PHP

Java脚本

输出： 

1
6