PAT 甲级 1152 Google Recruitment (20分)(简单题，直接模拟)

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

The natural constant

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 ( $\leq 1,000) and K (< 10), which are the numbers of digits of the given number and the prime to be found, respectively. Then the L-digit number N is given in the next line.$

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:

题意：给定n和k，n表示要输入的数据的长度，k表示在输入的数据中找出长度为k的素数（第一次出现），找到输出这个素数，如果没有输出404。

解题思路：直接模拟，在主函数外写一个判断是否是素数的函数，输入的数据很大，所以用字符串处理，素数的个位基本上是奇数，除了2，但题目说了可以有前导0，所以我们还要考虑个位是2的情况，在字符串找到奇数，然后往前去k个，组成一个字符串，将其转化成整数，然后调用素数判断函数，找到输出这个字符串，找不着就输出404。
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原文链接：https://blog.csdn.net/Imagirl1/article/details/85330953

#include<bits/stdc++.h>
using namespace std;
char a[1005];
bool prime(int x){
    if(x<2) return false;
    if(x==2) return true;
    for(int i=2;i<sqrt(x);i++){
        if(x%i==0){
            return false;
        }
    }
    return true;
}
int main(){
    int n,k;
    cin>>n>>k;
    cin>>a;
    int l=strlen(a);
    int x=0;
    int f=0;
    for(int i=0;i<=l-k;i++){
        x=0;
        for(int j=i;j<=i+k-1;j++){
            x=x*10+a[j]-'0';
        }
        //cout<<x<<endl;
        if(prime(x)){
            for(int j=i;j<=i+k-1;j++){            
                cout<<a[j];
            }
            f=1;
            break;
        }
    }
    if(!f){
        cout<<"404";
    }
    return 0;
}