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Program to show decimal number from rational number representation in C++
Suppose we have two numbers called numerator and denominator representing a rational number in the form (numerator / denominator). We have to find its decimal representation as a string. If there are some recurring numbers, then surround them with brackets.
So, if the input is like numerator = 164 denominator = 3, then the output will be "54.(6)".
To solve this, we will follow these steps −
- if numerator is same as 0, then −
- return "0"
- Define an array ans
- if numerator < 0 and denominator > 0 or numerator > 0 and denominator < 0, then −
- insert '-' at the end of ans
- divisor := |numerator|
- dividend := |denominator|
- remainder := divisor mod dividend
- x := convert (divisor / dividend) to string
- for initialize i := 0, when i < size of x, update (increase i by 1), do −
- insert x[i] at the end of ans
- if remainder is same as 0, then −
- return ans as string
- insert '.' at the end of ans
- Define one map m
- while remainder is not equal to 0, do −
- if remainder is not in m, then −
- insert (first element of ans concatenate '(') into ans
- insert ')' at the end of ans
- Come out from the loop
- Otherwise −
- m[remainder] := size of ans
- remainder := remainder * 10
- insert (remainder / dividend) concatenate '0' at the end of ans
- remainder := remainder mod dividend
- if remainder is not in m, then −
- return ans as string
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string solve(int numerator, int denominator) {
if (numerator == 0)
return "0";
vector<char> ans;
if (numerator < 0 && denominator > 0 || numerator > 0 && denominator < 0)
ans.push_back('-');
long divisor = labs(numerator);
long dividend = labs(denominator);
long remainder = divisor % dividend;
string x = to_string(divisor / dividend);
for (int i = 0; i < x.size(); i++) {
ans.push_back(x[i]);
}
if (remainder == 0) {
return string(ans.begin(), ans.end());
}
ans.push_back('.');
map<int, int> m;
while (remainder != 0) {
if (m.find(remainder) != m.end()) {
ans.insert(ans.begin() + m[remainder], '(');
ans.push_back(')');
break;
} else {
m[remainder] = ans.size();
remainder *= 10;
ans.push_back((remainder / dividend) + '0');
remainder %= dividend;
}
}
return string(ans.begin(), ans.end());
}
};
string solve(int numerator, int denominator) {
return (new Solution())->solve(numerator, denominator);
}
int main() {
cout << solve(164, 3);
}Input
164, 3
Output
54.(6)
Updated on: 12-Dec-2020
281 Views
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