Question

我试图通过计算找到的素数来将我的结果与Wolfram Alpha的结果进行比较。

看起来效果很好，没有错误。

但是当我向SPOJ提交此解决方案时，它会显示错误消息“错误答案”。

我还尝试将最终打印的end =更改为''（空白字符串），但仍然得到“错误答案”。

不确定我的筛选算法是否有问题，或输出错误。

**编辑：问题http://www.spoj.pl/problems/PRIME1/的链接

这是我的PRIME1源代码，希望有人可以指出我的错。非常感谢。

希望有人可以教我如何为这样的程序编写测试，我还在学习，不知道如何对程序进行自动化测试，但想学习。

    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

这部分逻辑

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

错了。你的elif条件在这里没有多大意义。如果n不是minNum的除数，则n不小于minNum的最小倍数为nMultipleMostCloseToMin + n，无论sqrtOfMaxNum是否为minNum是否大于n <= minNum。你在这里想要的条件是{{1}}，以避免越过素数本身。

Python中SPOJ PRIME1的错误答案

1 个答案: