Given a fraction, find recurring sequence of digits if exists, otherwise print “No recurring sequence”.

Examples:

Input : Numerator = 8, Denominator = 3 Output : Recurring sequence is 6 Explanation : 8/3 = 2.66666666....... Input : Numerator = 50, Denominator = 22 Output : Recurring sequence is 27 Explanation : 50/22 = 2.272727272..... Input : Numerator = 11, Denominator = 2 Output : No recurring sequence Explanation : 11/2 = 5.5

## We strongly recommend that you click here and practice it, before moving on to the solution.

**When does the fractional part repeat ?**

Let us simulate the process of converting fraction to decimal. Let us look at the part where we have already figured out the integer part which is floor(numerator/denominator). Now we are left with ( remainder = numerator%denominator ) / denominator.

If you remember the process of converting to decimal, at each step we do the following :

- Multiply the remainder by 10.
- Append remainder / denominator to result.
- Remainder = remainder % denominator.

At any moment, if remainder becomes 0, we are done.

However, when there is a recurring sequence, remainder never becomes 0. For example if you look at 1/3, the remainder never becomes 0.

Below is one important observation :

If we start with remainder ‘rem’ and if the remainder repeats at any point of time, the digits between the two occurrence of ‘rem’ keep repeating.

So the idea is to store seen remainders in a map. Whenever a remainder repeats, we return the sequence before the next occurrence. Below is C++ implementation of above idea.

`// C++ program to find repeating sequence in a fraction ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// This function returns repeating sequence of ` `// a fraction. If repeating sequence doesn't ` `// exits, then returns empty string ` `string fractionToDecimal(` `int` `numr, ` `int` `denr) ` `{ ` ` ` `string res; ` `// Initialize result ` ` ` ` ` `// 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 <` `int` `, ` `int` `> mp; ` ` ` `mp.clear(); ` ` ` ` ` `// Find first remainder ` ` ` `int` `rem = numr%denr; ` ` ` ` ` `// 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 / denr; ` ` ` `res += to_string(res_part); ` ` ` ` ` `// Update remainder ` ` ` `rem = rem % denr; ` ` ` `} ` ` ` ` ` `return` `(rem == 0)? ` `""` `: res.substr(mp[rem]); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `numr = 50, denr = 22; ` ` ` `string res = fractionToDecimal(numr, denr); ` ` ` `if` `(res == ` `""` `) ` ` ` `cout << ` `"No recurring sequence"` `; ` ` ` `else` ` ` `cout << ` `"Recurring sequence is "` `<< res; ` ` ` `return` `0; ` `} ` |

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Output :

Recurring sequence is 27

This article is contributed by **Dhruv Mahajan**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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