PAT日志 1152

顽强的小白

1152 Google Recruitment （20 分）

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:

49877

Sample Input 2:

10 3
2468024680

Sample Output 2:

404

题目解析

给出一个超长的数字，给出一个位数，输出数字中给定位数的素数，没有就输出404。
题目简单，我还用了一个不常用的函数string中的substr()。
substr(pos,len)：得到从pos位置开始长度位len的字符串。完全是为这道题量身定做的四不四。

注意点：

• 首先是循环取子串的时候，最后可取的位置不是到末尾，而是由字串的长度决定的。
• 最后如果串的长度小于字串要求的长度，直接输出404，不然4号case过不去。

代码实现

#include <cstdio> 
#include <iostream> 
#include <string> 
#include <cmath> 
using namespace std; 
bool judge(string s){
 int num=0;
 for(int i=0;i<s.size();++i){
  num*=10;
  num+=s[i]-'0';
 }
 for(int i=2;i<=sqrt(num);++i){
  if(num%i==0){
   return false;
  }
 }
 return true;
}
int main(){
 int n,k;
 scanf("%d%d",&n,&k);
 if(n<k) {
  printf("404\n");
  return 0;
 }
 string s,son;
 cin>>s;
 int i;
 for(i=0;i<=s.size()-k;++i){
  son=s.substr(i,k);
  if(judge(son)==true){
   cout<<son<<endl;
   return 0;
  }
 }
 if(i>s.size()-k) printf("404\n");
 return 0;
}