How to use oneleft method in pyatom

Best Python code snippet using pyatom_python

PowerBinaryCalculator.py

Source:PowerBinaryCalculator.py Github

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1#generate inputs 2input = open('input.txt','r')3numbers = input.read().splitlines()4cNum= numbers.copy()5#values used in script6lenLine=len(numbers[0])7sigBit,oneLeft="0",false8#calculate oxygen9#for every bit position(0->11 in this case)10for bitPos in range(0,lenLine):11 #count the number of 0s and 1s 12 zeros,ones=0,013 for i in range(0,len(numbers)):14 if((numbers[i])[bitPos]=="0"):15 zeros+=116 else:17 ones+=118 #if ones>=zeros:19 if(ones>=zeros):20 sigBit="1"21 else:22 sigBit="0"23 #remove any items in the list that don't meet oxygen specifications24 i=025 while(i<len(numbers)):26 if(len(oxNum)==1):27 oneLeft=true28 break29 if((numbers[i])[bitPos]!=sigBit):30 numbers.pop(i)31 else:32 i+=133 if(oneLeft): #stop if only one result left34 break35#calculate c0236#for every bit position(0->11 in this case)37oneLeft=false38for bitPos in range(0,lenLine):39 #count the number of 0s and 1s 40 zeros,ones=0,041 for i in range(0,len(cNum)):42 if((cNum[i])[bitPos]=="0"):43 zeros+=144 else:45 ones+=146 #if ones>=zeros:(inverse of oxygen)47 if(ones>=zeros):48 sigBit="0"49 else:50 sigBit="1"51 #remove any items in the list that don't meet oxygen specifications52 i=053 while(i<len(cNum)):54 if(len(cNum)==1):55 oneLeft=true56 break57 if((cNum[i])[bitPos]!=sigBit):58 cNum.pop(i)59 else:60 i+=161 if(oneLeft): #stop if only one result left62 break63#generate and print results64oxygen=int(numbers[0],2)65co2=int(cNum[0],2)...

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postfixlogic.py

Source:postfixlogic.py Github

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1__author__ = "Dmitry Philippov"2import sys3sys.stdout = open("postfixlogic.out", "w")4def main():5 print("""16S 0 -> now 0 >7S 1 -> now 1 >8now _ -> fillblank _ <9fillblank * -> fillblank _ <10fillblank 0 -> AC 0 ^11fillblank 1 -> AC 1 ^12now 0 -> zeroleft * <13zeroleft _ -> setzero _ >14zeroleft * -> zeroleft * <15zeroleft _ -> now _ ^16zeroleft 0 -> setzero 0 >17zeroleft 1 -> setzero 1 >18setzero * -> zeroright 0 >19zeroright * -> zeroright * >20zeroright _ -> now _ ^21zeroright 0 -> now 0 ^22zeroright 1 -> now 1 ^23zeroright o -> now o ^24zeroright a -> now a ^25now 1 -> oneleft * <26oneleft _ -> setone _ >27oneleft * -> oneleft * <28oneleft 0 -> setone 0 >29oneleft 1 -> setone 1 >30setone * -> oneright 1 >31oneright * -> oneright * >32oneright 0 -> now 0 ^33oneright 1 -> now 1 ^34oneright o -> now o ^35oneright a -> now a ^36now a -> andleft * <37andleft * -> andleft * <38andleft 0 -> andleftzero * <39andleftzero 0 -> andright 0 >40andleftzero 1 -> andright 0 >41andright * -> andright * >42andright _ -> now _ ^43andright 0 -> now 0 ^44andright 1 -> now 1 ^45andright o -> now o ^46andright a -> now a ^47andleft 1 -> andleftone * <48andleftone 0 -> andright 0 >49andleftone 1 -> andright 1 >50now o -> orleft * <51orleft * -> orleft * <52orleft 0 -> orleftzero * <53orleftzero 0 -> orright 0 >54orleftzero 1 -> orright 1 >55orright * -> orright * >56orright _ -> now _ ^57orright 0 -> now 0 ^58orright 1 -> now 1 ^59orright o -> now o ^60orright a -> now a ^61orleft 1 -> orleftone * <62orleftone 0 -> orright 1 >63orleftone 1 -> orright 1 >""")64if __name__ == "__main__":...

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034-median-of-two-sorted-arrays.py

Source:034-median-of-two-sorted-arrays.py Github

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1class Solution:2 def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:3 m, n = len(nums1), len(nums2)4 halfLen = (m + n) // 25 6 if m > n:7 nums1, nums2 = nums2, nums18 m, n = n, m9 10 l, r = 0, m - 111 while True:12 i = l + (r - l) // 213 j = halfLen - i - 214 15 oneLeft = nums1[i] if i >= 0 else float('-infinity')16 oneRight = nums1[i + 1] if (i + 1) < m else float('infinity')17 twoLeft = nums2[j] if j >= 0 else float('-infinity')18 twoRight = nums2[j + 1] if (j + 1) < n else float('infinity')19 20 if oneLeft <= twoRight and twoLeft <= oneRight:21 if (m + n) % 2:22 return min(oneRight, twoRight)23 return (min(oneRight, twoRight) + max(oneLeft, twoLeft)) / 224 elif oneLeft > twoRight:25 r = i - 126 else:...

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