根据Midy定理，如果重复小数点的周期为
，其中p是素数， 
是一个减少的分数，具有偶数个数字，然后将重复部分分成两半，相加得到字符串9s。
例子 ：
a = 1 and p = 7
1/7 = 0.14285714285..
So 1/7 is a repeating decimal with 142857 being repeated. Now, according to the theorem, it has even number of repeating digits i.e. 142857. Further if we divide this into two halves, we get 142 and 857. Thus, on adding these two, we get 999 which is string of 9s and matches our theorem.
a = 2 and p = 11
2/11 = 0.18181818181..
Here, repeating decimal is 18. Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.
给定分子和分母，任务是查找结果浮点数是否遵循Midy定理。
方法 ：
让我们模拟一下将分数转换为十进制的过程。让我们看一下已经计算出整数部分的那部分，即楼板(分子/分母)。现在我们剩下(余数=分子％分母)/分母。
如果您还记得转换为十进制的过程，则在每个步骤中，请执行以下操作：
- 余数乘以10。
- 追加余数/分母以得出结果。
- 余数=余数％分母。
在任何时候，如果余数变为0，我们就完成了。
但是，当存在重复序列时，余数永远不会变为0。例如，如果您查看1/3，则余数永远不会变为0。
以下是一项重要观察：
如果我们以余数’rem’开头，并且余数在任何时间点重复，那么两次’rem’出现之间的数字将继续重复。
因此，想法是将可见的剩余物存储在地图中。每当余数重复时，我们将在下一次出现之前返回序列。
以下是Midy定理的实现：
C++
// C++ implementation as a
// proof of the Midy's theorem
#include
using namespace std;
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
string fractionToDecimal(int numerator, int denominator)
{
string res;
/* Create a map to store already seen remainders
remainder is used as key and its position in
result is stored as value. Note that we need
position for cases like 1/6. In this case,
the recurring sequence doesn't start from first
remainder. */
map mp;
mp.clear();
// Find first remainder
int rem = numerator % denominator;
// Keep finding remainder until either remainder
// becomes 0 or repeats
while ((rem != 0) && (mp.find(rem) == mp.end()))
{
// Store this remainder
mp[rem] = res.length();
// Multiply remainder with 10
rem = rem * 10;
// Append rem / denr to result
int res_part = rem / denominator;
res += to_string(res_part);
// Update remainder
rem = rem % denominator;
}
return (rem == 0) ? "-1" : res.substr(mp[rem]);
}
// Checks whether a number is prime or not
bool isPrime(int n)
{
for (int i = 2; i <= n / 2; i++)
if (n % i == 0)
return false;
return true;
}
// If all conditions are met,
// it proves Midy's theorem
void Midys(string str, int n)
{
int l = str.length();
int part1 = 0, part2 = 0;
if (!isPrime(n))
{
cout << "Denominator is not prime, "
<< "thus Midy's theorem is not applicable";
}
else if (l % 2 == 0)
{
for (int i = 0; i < l / 2; i++)
{
part1 = part1 * 10 + (str[i] - '0');
part2 = part2 * 10 + (str[l / 2 + i] - '0');
}
cout << part1 << " + " << part2 << " = "
<< (part1 + part2) << endl;
cout << "Midy's theorem holds!";
}
else
{
cout << "The repeating decimal is of odd length "
<< "thus Midy's theorem is not applicable";
}
}
// Driver code
int main()
{
int numr = 2, denr = 11;
string res = fractionToDecimal(numr, denr);
if (res == "-1")
cout << "The fraction does not have repeating decimal";
else {
cout << "Repeating decimal = " << res << endl;
Midys(res, denr);
}
return 0;
} Java
// Java implementation as a
// proof of the Midy's theorem
import java.util.*;
class GFG{
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
static String fractionToDecimal(int numerator,
int denominator)
{
String res = "";
/* Create a map to store already seen remainders
remainder is used as key and its position in
result is stored as value. Note that we need
position for cases like 1/6. In this case,
the recurring sequence doesn't start from first
remainder. */
HashMap mp = new HashMap<>();
// Find first remainder
int rem = numerator % denominator;
// Keep finding remainder until either remainder
// becomes 0 or repeats
while ((rem != 0) && !mp.containsKey(rem))
{
// Store this remainder
mp.put(rem, res.length());
// Multiply remainder with 10
rem = rem * 10;
// Append rem / denr to result
int res_part = rem / denominator;
res += res_part + "";
// Update remainder
rem = rem % denominator;
}
return (rem == 0) ? "-1" : res.substring(mp.get(rem));
}
// Checks whether a number is prime or not
static boolean isPrime(int n)
{
for(int i = 2; i <= n / 2; i++)
if (n % i == 0)
return false;
return true;
}
// If all conditions are met,
// it proves Midy's theorem
static void Midys(String str, int n)
{
int l = str.length();
int part1 = 0, part2 = 0;
if (!isPrime(n))
{
System.out.print("Denominator is not prime, " +
"thus Midy's theorem is not " +
"applicable");
}
else if (l % 2 == 0)
{
for(int i = 0; i < l / 2; i++)
{
part1 = part1 * 10 + (str.charAt(i) - '0');
part2 = part2 * 10 + (str.charAt(l / 2 + i) - '0');
}
System.out.println(part1 + " + " + part2 +
" = " + (part1 + part2));
System.out.print("Midy's theorem holds!");
}
else
{
System.out.print("The repeating decimal is " +
"of odd length thus Midy's "+
"theorem is not applicable");
}
}
// Driver code
public static void main(String []args)
{
int numr = 2, denr = 11;
String res = fractionToDecimal(numr, denr);
if (res == "-1")
System.out.print("The fraction does not " +
"have repeating decimal");
else
{
System.out.println("Repeating decimal = " + res);
Midys(res, denr);
}
}
}
// This code is contributed by rutvik_56 C#
// C# implementation as a
// proof of the Midy's theorem
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
static String fractionToDecimal(int numerator,
int denominator)
{
String res = "";
/* Create a map to store already seen remainders
remainder is used as key and its position in
result is stored as value. Note that we need
position for cases like 1/6. In this case,
the recurring sequence doesn't start from first
remainder. */
Dictionary mp = new Dictionary();
// Find first remainder
int rem = numerator % denominator;
// Keep finding remainder until either remainder
// becomes 0 or repeats
while ((rem != 0) && !mp.ContainsKey(rem))
{
// Store this remainder
mp[rem]= res.Length;
// Multiply remainder with 10
rem = rem * 10;
// Append rem / denr to result
int res_part = rem / denominator;
res += res_part + "";
// Update remainder
rem = rem % denominator;
}
return (rem == 0) ? "-1" : res.Substring(mp[rem]);
}
// Checks whether a number is prime or not
static bool isPrime(int n)
{
for(int i = 2; i <= n / 2; i++)
if (n % i == 0)
return false;
return true;
}
// If all conditions are met,
// it proves Midy's theorem
static void Midys(String str, int n)
{
int l = str.Length;
int part1 = 0, part2 = 0;
if (!isPrime(n))
{
Console.Write("Denominator is not prime, " +
"thus Midy's theorem is not " +
"applicable");
}
else if (l % 2 == 0)
{
for(int i = 0; i < l / 2; i++)
{
part1 = part1 * 10 + (str[i] - '0');
part2 = part2 * 10 + (str[l / 2 + i] - '0');
}
Console.WriteLine(part1 + " + " + part2 +
" = " + (part1 + part2));
Console.Write("Midy's theorem holds!");
}
else
{
Console.Write("The repeating decimal is " +
"of odd length thus Midy's "+
"theorem is not applicable");
}
}
// Driver code
public static void Main(string []args)
{
int numr = 2, denr = 11;
string res = fractionToDecimal(numr, denr);
if (res == "-1")
Console.Write("The fraction does not " +
"have repeating decimal");
else
{
Console.WriteLine("Repeating decimal = " + res);
Midys(res, denr);
}
}
}
// This code is contributed by pratham76. 输出 ：
Repeating decimal = 18
1 + 8 = 9
Midy's theorem holds!有关Midy定理的更多信息，请参见
http://www.kurims.kyoto-u.ac.jp/EMIS/journals/INTEGERS/papers/h2/h2.pdf
http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1047&context=mathfacpub