decimal to fraction calculator

Leetcode: Fraction to Recurring Decimal, leetcoderecurring Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. For example, Given numerator = 1, denominator = 2, return "0.5

Given integers representing the numerator and denominator of a fraction, return the fraction in string format.If the fractional part was repeating, enclose the repeating part in parentheses.For example,Given numerator = 1, denominator = 2, return "0.5".Given numerator = 2, denominator = 1, return "2".Given numerator = 2, denominator = 3, return "0. (6) ".Problem Solving Ideas:A variety of critical values wi

Title Description:Given integers representing the numerator and denominator of a fraction, return the fraction in string format.If the fractional part was repeating, enclose the repeating part in parentheses.For example,Given numerator = 1, denominator = 2, return "0.5".Given numerator = 2, denominator = 1, return "2".Given numerator = 2, denominator = 3, return "0. (6) ".is to give a divisor and dividend a

Given integers representing the numerator and denominator of a fraction, return the fraction in string format.If the fractional part was repeating, enclose the repeating part in parentheses.Example 1:Input:numerator = 1, denominator = 2Output: "0.5"Example 2:Input:numerator = 2, denominator = 1Output: "2"Example 3:Input:numerator = 2, denominator = 3Output: "0. (6)"Class Solution:def Fractiontodecimal (self

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. For example, Given Numerator = 1, Denominator = 2, return "0.5 ".Given Numerator = 2, Denominator = 1, return "2 ".Given Numerator = 2, Denominator = 3, return "0. (6 )".Credits:Special thanks to @ Shangrila for add

(); BooleanIsnegative=false; if(Numerator ) {isnegative= !isnegative; } if(denominator) {isnegative= !isnegative; } if(isnegative) {res.append (‘-‘); } LongReminder = num%den; Res.append (Num/den); intSize =res.length (); BooleanHascycle =false; while(Reminder! = 0 ){ if(Container.isempty ()) {Res.append (‘.‘); Size++; } if(!Container.containskey (Reminder)) {Size++; LongNewquoient = reminder*10/den; Res.append (

avoid the problem in 1, so the error value in this way.Case that might hang:0/111/90;1/2147483647 (Integer.max_value);-1/2147483647;1/-2147483638;The code is as follows: PublicString Fractiontodecimal (intA1,intA2) { if(A1 = = 0 | | a2 = = 0)return"0"; StringBuilder SB=NewStringBuilder (); Sb.append ((A1>0) ^ (a2>0)? "-":""); LongA = Math.Abs ((Long) A1); Longb = Math.Abs ((Long) A2); Sb.append (Math.Abs (a/b)); A= (a% b) * 10; if(A! = 0) Sb.append ("."); MapNewHashmap(); Map.put (NewLon

Given integers representing the numerator and denominator of a fraction, return the fraction in string format.If the fractional part was repeating, enclose the repeating part in parentheses.For example, Given numerator = 1, denominator = 2, return "0.5". Given numerator = 2, denominator = 1, return "2". Given numerator = 2, denominator = 3, return "0. (6) ". Credits:Special thanks to @

result at - //after you determine the symbol, you can convert two numbers to integer operations, but for negative (1 - LL N; - LL m; -n=numerator0?-Numerator:numerator; -m=denominator0?-Denominator:denominator; in - ints=-1;//flag Cycle section start position to + -LL intpart=n/m; then=n%m; *Ret+=itos (Intpart);//Integer Part $ if(n==0)returnret;Panax Notoginseng - inttag=1;//The tag indicates whether the Loop section was found, when the tag==1 indic

Note that there are many places to deal with long and intsuch as Assignment N, d to determine if it should be a negative number.If writtenlong n = (long) numerator>0? Numerator:-numerator; long D = (long) denominator>0? Denominator:-denominator;There will be an error, possibly a numerator as int too short, so the front plus minus sign error Public classSolution { PublicString Fractiontodecimal (intNumerator,intdenominator) { //I find it particularly clear to benefit from a solution to htt

Ideas: 1, the molecular 0 can be returned in advance2, the result is positive and negative judgment3. Array out of bounds (minimum negative number divided by -1)4, for the beginning of the encounter decimal to add ".", there is a decimal can expand dividend, for whether there is a repetition of the place to judge whether its dividend is repeated!!!For the ABS () function, remember to convert the number to a

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=1717Analysis:We divide this repeating decimal into three parts 0.a1a2a3a4....an (B1B2BB3...BM) into 0, non-cyclic part a1a2...an, and loop section B1B2B3...BMSet this decimal for X,A1A2A3....AN=S1, b1b2b3...bm=s2,y=0. (B1B2B3B4. BmX * 10^n = s1 + y; ---------1)Y * 10^m = s2 + y; ----------2)by 2) y = s2/(10^m-1); ------------3)1), 3) Unitedx= (s2 + s1 *

) - return "0"; +mapLong,Long>Mymap; -vectorLong>Shangvec; + LongYup =0; A LongFirstflag =1; at while(bei) - { - if(Firstflag = =0) -Bei *=Ten; -Firstflag =0; - LongTMP = bei/Chu; in LongYu = bei-tmp *Chu; - Shangvec.push_back (TMP); tomapLong,Long>:: iterator it; +it =Mymap.find (Yu); - if(it!=mymap.end ()) the { * LongXUNP = it->second; $ stringAns ="";Panax

various situations. There are disgusting negative numbers the most value, with a long long to do. In addition to the situation below are listed here./*1, 8 = 0.1251, 6 = 0.1 (6) -50, 6 = -6.250,-3 = 0-1, -2147483648 = "0.0000000004656612873077392578125" */typedef long Long Llong;class Solution {public:string fractiontodecimal (int numerator, int denominator) {if (numerator = = 0) { return "0"; The string result; Llong n = numerator; Llong d = denominator;

The result is a loop after the decimal point, with parentheses enclosing the part of the loop that appears.Find out the loop part of the idea:Maintains a unordered_mapThe idea of division:Remain stores the remainder of each division, while (remain) loop, each remain *= in the loop, res + = Remain/denominator, remain = remain% denominator.1 classSolution {2 Public:3 stringFractiontodecimal (intNumerator,intdenominator) {4 if(!numerator)ret

Notes:1. When numerator is 0, return "0". Check This corner case, because 0/-5 would return-0.2. Use a long long int for DIVD and divs, mainly for divs. Because when divs = Int_min, Fabs (divs) is large.3. Forget to rest *= to get the next level.1 classSolution {2 Public:3 stringFractiontodecimal (intNumerator,intdenominator) {4 if(Numerator = =0)return "0";5 stringresult;6 if(Numerator 0^ Denominator 0) result ="-";7 Long Long intDIVD = Fabs (numerator), div

One of the count, if there is a repetition of the remainder of the cycle is the beginning of the festival.Use Hashtable to record the position and insert parentheses.Note negative numbers.classSolution { Public: stringFractiontodecimal (intNumerator,intdenominator) { intSIGN1 = Numerator >=0?1: -1; intSIGN2 = Denominator >=0?1: -1; Long Longnum = (Long Long) numerator; Long LongDen = (Long Long) denominator; Num=ABS (num); Den=abs (DEN); Long Longd = num/den; Long Longrem = num%den; Un