题目

In July 2004, Google posted on a giant billboard along Highway 101 in Silicon Valley (shown in the picture below) for recruitment. The content is super-simple, a URL consisting of the first 10-digit prime found in consecutive digits of the natural constant e. The person who could find this prime number could go to the next step in Google’s hiring process by visiting this website.

prime

The natural constant e is a well known transcendental number（超越数）. The first several digits are: e = 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921… where the 10 digits in bold are the answer to Google’s question.

Now you are asked to solve a more general problem: find the first K-digit prime in consecutive digits of any given L-digit number.

Input Specification:

Each input file contains one test case. Each case first gives in a line two positive integers: L (≤ 1,000) and K (< 10), which are the numbers of digits of the given number and the prime to be found, respectively. Then the L-digit number N is given in the next line.

Output Specification:

For each test case, print in a line the first K-digit prime in consecutive digits of N. If such a number does not exist, output 404 instead. Note: the leading zeroes must also be counted as part of the K digits. For example, to find the 4-digit prime in 200236, 0023 is a solution. However the first digit 2 must not be treated as a solution 0002 since the leading zeroes are not in the original number.

Sample Input 1:

1
2
20 5
23654987725541023819

Sample Output 1:

1
49877

Sample Input 2:

1
2
10 3
2468024680

Sample Output 2:

1
404

题解

思路

  • 给定一场串数和一个长度，让你在这串数中找到一个字符串，组成的数字是素数，字符串的长度刚好等于给定的。有多个就排在第一个的。
  • 那么就以这个长度为窗口，不断滑动，即可。
  • 注意输出的时候要输出切分的字符串，不能输出整数，不然前面的0会被吃掉。

数据结构

  • num 是原数
  • test 是切分后的字符串

算法

  • 求素数的算法
    • 先判断123
    • 再判断偶数
    • 再从3到sqrt(num)，每隔2个递增，依次判断这个数能不能被遍历的数整除

代码

  • 由于使用Python能够AC，因此只放了Python的题解。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
def isPrime(a):
    if a == 1:
        return False
    if a == 2 or a == 3:
        return True
    if a % 2 == 0:
        return False
    for i in range(3, int(a ** 0.5) + 1, 2):
        if a % i == 0:
            return False
    return True


length, digit = list(map(int, input().split()))
num = input()
for i in range(len(num) - digit + 1):
    test = num[i:i + digit]
    if isPrime(int(test)):
        print(test)
        break
else:
    print("404")