PAT甲级——A1152 GoogleRecruitment【20】

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.

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:

20 5

23654987725541023819

Sample Output 1:

Sample Input 2:

10 3

2468024680

Sample Output 2:

404

Solution：
　　这道题令我惊讶的是，竟然是简单的判断一下是不是素数？！
　　本以为这么大的数字级别，应该是建立素数表来判断的，想不到竟然时一个个数字进行简单的判断是不是素数？
　　倒是建立素数表内存超了，判断是不是素数竟然没有超时？！
　　下面代码给出了建立素数表

 #include <iostream>

 #include <vector>

 #include <string>

 #include <cmath>

 using namespace std;

 int n, k;

 string str, res = "";

 //void getPrimeTable(int inf, vector<bool>&notPrime)//创建素数表

 //{

 //    notPrime[0] = notPrime[1] = true;

 //    for (int i = 2; i <= inf; ++i)

 //        if (notPrime[i] == false)//从2这个素数开始

 //            for (int j = 2; j*i <= inf; ++j)

 //                notPrime[j*i] = true;//剔除素数的所有倍数

 //}

 bool isPrime(int x)//判断是不是素数

 {

     if (x < )

         return true;

     for (int i = ; i*i <= x; ++i)

         if (x%i == )

             return false;

     return true;

 }

 int main()

 {

     cin >> n >> k;

     cin >> str;

     //int size = (int)pow(10, k);

     //vector<bool>notPrime(size+1, false);//防止内存太大，我这里是动态建立数组的

     //getPrimeTable(size, notPrime);//创建素数表

     for (int i = ; i + k <= n; ++i)

     {

         string s = str.substr(i, k);

         int num = atoi(s.c_str());

         if (isPrime(num))//notPrime[num]==false)//使用的代码简单的素数判断，注释的代码是使用素数表

         {

             res = s;

             break;

         }

     }

     if (res.size() > )

         cout<<res;

     else

         cout << "";

     return ;

 }