Cycle Length

Search BackLinks only
Display context of search results
Case-sensitive searching

• 3N+1Problem/Leonardong . . . . 18 matches
        def getMaximumCycleLength(self, aFrom, aTo):
        cycleLength = self.getCycleLength(i)
        if max < cycleLength:
        max = cycleLength
        def getCycleLength( self, aStart ):
        cycleLength = 1
        cycleLength = cycleLength + 1
        return cycleLength
        print self.getMaximumCycleLength(numFrom, numTo)
        def testGetMaximumCycleLength(self):
        self.assertEquals( 1, self.runner.getMaximumCycleLength( 1, 1 ) )
        self.assertEquals( 20, self.runner.getMaximumCycleLength( 1, 10 ) )
        self.assertEquals( 125, self.runner.getMaximumCycleLength( 100, 200 ) )
        self.assertEquals( 89, self.runner.getMaximumCycleLength( 201, 210 ) )
        self.assertEquals( 174, self.runner.getMaximumCycleLength( 900, 1000) )
        def testGetCycleLength(self):
        self.assertEquals( 1, self.runner.getCycleLength(1) )
        self.assertEquals( 5, self.runner.getCycleLength(16) )
        self.assertEquals( 16, self.runner.getCycleLength(22) )
        * MAX(100000)개의 원소를 가진 리스트에 계산했던 CycleLength를 저장한다.
• 3N+1Problem/1002_2 . . . . 10 matches
       class CycleLength:
        def maxCycleLengthInRange(self,i,j):
        >>> c=CycleLength()
        >>> c.maxCycleLengthInRange(1,10)
        >>> c.maxCycleLengthInRange(100,200)
        >>> c.maxCycleLengthInRange(201,210)
        >>> c.maxCycleLengthInRange(900,1000)
        c=CycleLength()
        print c.maxCycleLengthInRange(1,999999)
        ''{{{~cpp CycleLength.value}}} 와 거의 비슷한 의사코드가 [이덕준]의 연습장에도... 무척 반가움. --[이덕준]''
• 3N+1Problem/강희경 . . . . 8 matches
       def FindMaxCycleLength(aMin, aMax, aBinaryMap):
        maxCycleLength = 0
        cycleLength = TreeNPlusOne(i, aMin, aMax, aBinaryMap)
        if(maxCycleLength < cycleLength):
        maxCycleLength = cycleLength
        return maxCycleLength
       def OutputResult(aMin, aMax, aMaxCycleLength):
        print aMin, aMax, aMaxCycleLength
        OutputResult(min, max, FindMaxCycleLength(min, max, binaryMap))
• 3N+1/임인택 . . . . 5 matches
        mergeList numbers (map maxCycleLength numbers) []
       maxCycleLength fromto =
        head (List.sortBy (flip compare) (gatherCycleLength (head fromto) (head (tail fromto)) []) )
       gatherCycleLength num to gathered =
        else gatherCycleLength (num+1) to ( gathered ++ [doCycle num 1])

Found 4 matching pages out of 7540 total pages (5000 pages are searched)

You can also click here to search title.