Question

I’ve tried to compare my result to Wolfram Alpha’s result by counting primes found.

It seems works well, without error.

But when I submit this solution to SPOJ, it shows the error message “wrong answer”.

I also tried to change the final print’s end= to ” (blank string), but still got “wrong answer”.

Not sure if is there something wrong with my sieve algorithm, or output error.

**edit: link of problem http://www.spoj.pl/problems/PRIME1/

Here is my PRIME1 source code, hope someone can point out my fault. Thanks a lot.

And hope someone can teach me how to write a test to programs like this, I’m still learning, don’t know how to do automated tests to programs, but want to learn.

    def getPrimeInRange(minNum, maxNum):
        #just a variation with skipping step of the  Sieve of E's
        processingRange = list(range(minNum, maxNum+1))
        #prefix, due to 1 is not a prime
        if minNum == 1:
            processingRange[0] = 0
        sqrtOfMaxNum = int(maxNum ** 0.5) + 1
        primesUnderSqrt = list(range(sqrtOfMaxNum))
        #prefix, due to 1 is not a prime
        primesUnderSqrt[1] = 0

        #my strategy is flip all numbers that is not a prime to zero, which equals to False.
        #for Sieve of E's, all the primes under sqrt of max num are the only needed numbers to sieve primes out.
        #so here is a smaller Sieve of E's for numbers under sqrt
        for n in primesUnderSqrt:
            if n: #which equals to "if n != 0"
                nowIndex = n + n
                while True:
                    try:
                        primesUnderSqrt[nowIndex] = 0
                        nowIndex += n
                    except IndexError:
                        break

        #full aspect sieve
        for n in primesUnderSqrt:
            if n:
                #for easier flip processing, I need the offset number for the flipping.
                nMultipleMostCloseToMin = n * (minNum // n)
                if nMultipleMostCloseToMin == minNum:
                    nowIndex = 0
                elif sqrtOfMaxNum <= minNum:
                    nowIndex = nMultipleMostCloseToMin + n - minNum
                elif sqrtOfMaxNum > minNum:
                    nowIndex = nMultipleMostCloseToMin + n - minNum + n

                #happy flippin'
                while True:
                    try:
                        processingRange[nowIndex] = 0
                        nowIndex += n
                    except IndexError:
                        break

        return processingRange

    def main():
        todoTimes = int(input())
        todoNums = list(range(todoTimes))
        stringOutput = ''
        for i in range(todoTimes):
            todoNums[i] = input().split()
            todoNums[i][0] = int(todoNums[i][0])
            todoNums[i][1] = int(todoNums[i][1])
        for [minNum, maxNum] in todoNums:
            #countedNum = 0 #for algo debugging
            for n in getPrimeInRange(minNum, maxNum):
                if n:
                    stringOutput += str(n)
                    #countedNum += 1 #for algo debugging
                    stringOutput += '\n'
            stringOutput += '\n'
            #stringOutput += str(countedNum) #for algo debugging
        stringOutput = stringOutput.rstrip('\n')
        print(stringOutput)


    ifMainSucceed = main()

Answer 1

This part of your logic

if nMultipleMostCloseToMin == minNum:
    nowIndex = 0
elif sqrtOfMaxNum <= minNum:
    nowIndex = nMultipleMostCloseToMin + n - minNum
elif sqrtOfMaxNum > minNum:
    nowIndex = nMultipleMostCloseToMin + n - minNum + n

is wrong. Your elif-conditions don’t make much sense here. If n is not a divisor of minNum, the smallest multiple of n not less than minNum is nMultipleMostCloseToMin + n, regardless of whether sqrtOfMaxNum is larger than minNum or not. The condition you intended here was n <= minNum, to avoid crossing off the prime itself.