python fraction to decimal

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) ". Hint: No scary Mat

the corresponding left of every digit, assemble the result string in place.1 Public classSolution {2 PublicString Fractiontodecimal (LongNumerator,Longdenominator) {3 if(denominator==0)return"";4 if(numerator==0)return"0";5String res = "";6 if(numerator)7res = "-"; 8 9Numerator =Math.Abs (numerator);TenDenominator =Math.Abs (denominator); One A LongIntpart = numerator/denominator; - Longleft = numerator%denominator; -res =Res.concat (long.tostring

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.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 *

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

) - 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;

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

This article mainly introduces the implementation of the Python decimal decimal and binary fractional conversion function, combined with specific examples of the binary and decimal conversion of the principle and Python related implementation skills, the need for friends can

Python determines whether the instance is a positive decimal or a positive integer, and python decimal Python determines whether an instance is a positive decimal or a positive integer. Implementation Code: Def check_float (str

(' 0.3333 ')Fractions>>> from fractions import fraction>>> fraction ( -10) #分子分母Fraction (-8, 5)>>> fraction (123) #分子Fraction (123, 1)>>> fraction (' 3/7 ') #字符串分数Fraction (3, 7)>>>