欧拉计划问题编号 4

Question

n = 0
for a in xrange(999, 100, -1):
    for b in xrange(a, 100, -1):
        x = a * b
        if x > n:
            s = str(a * b)
            if s == s[::-1]:
                 n = a * b
print n

I have a question about this solution to the problem . I know it is right but I am wondering why in the xrange(999,100,-1) the -1 is there For the a and b for loops. Please explain. I am new to this :)

Answer 1

The third parameter to xrange() is the increment value. The default is 1, which means the counter will count in an increasing direction. To count in a decreasing direction, use -1. Your a counter will go from 999 to 101 (the xrange() iterator stops just before it reaches the second parameter value).

For future reference, see the xrange() documentation .

Answer 2

The -1 specifies a negative step. Thus moving from 999 descending to 100 (exclusive).

Answer 3

xrange function takes three arguments: start , stop and step . It returns a range of numbers starting from start continuing to stop , but not including it. If 'start' is bigger than stop , negative step must be provided.

So basically xrange(999, 100, -1) will give you [999, 998, ..., 101]

Answer 4

这意味着您正在以 1 为增量减少（因此是负号），从 999 到 100

Answer 5

This is what I got for the project Euler #4 question:

def Palindrome(s):
    if s == s[::-1]:
        return True
    else:
        return False

i = 100
j = 100
greatest = 0
while (i <= 999):
    while (j <= 999):
        product = i * j
        if (product > greatest and Palindrome(str(product))):
            greatest = product
        j += 1
    j = 100
    i += 1
print "Answer: " + str(greatest)

-M1K3

Answer 6

My Python solution:

container = []

for i in range(100, 999):
    for j in range(100, 999):
        num = i * j
        if str(num) == str(num)[::-1]:
            container.append(num)

print(max(container))

Answer 7

__author__ = 'shreysatapara'
f=0
for a in range(100,1000):
    for b in range(100,1000):
        c=a*b
        d=c
        s=0
        x=0
        for i in range(0,len(str(c))):
            s=c%10
            x=x*10+s
            c=c//10
            if(x==d):
                if(x>f):
                    f=x
                    print(f)

bingo answer is your last number in compiler....

Answer 8

import time
start_time = time.time()
largest_number = 0
for i in range(999,100,-1):
    for j in range(i,100,-1):
        string_number = str(i * j)
        if string_number == string_number[::-1]:
            if i*j > largest_number:
                largest_number = i * j
print(largest_number)
print(time.time() - start_time," seconds")

Answer 9

for x in range(100, 1000):
    for y in range(100, 1000):
        product = x*y
        if product > x*y:
            break
        if str(product) == str(product)[::-1] and product > 900000:
            print (x, y, product)

Answer 10

The negative one is the modifier. Basically the loops start at 999, end at 100, and get there by modifying each number by negative one.

Answer 11

def reverse(s):
    str = ""
    for i in s:
        str = i + str
    return str


MAXN = 0
for i in range(100, 1000):
    for j in range(100, 1000):
        a = i * j
        b = str(a)
        if b == reverse(b) and a > MAXN:
            MAXN = a
print(MAXN)