PAT 1152 Google Recruitmen (水题细心)

PAT 1152 Google Recruitmen

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.
PAT 1152 Google Recruitmen (水题细心)
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

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

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.

Examples

Sample Input 1:
20 5
23654987725541023819
Sample Output 1:
49877
Sample Input 2:
10 3
2468024680
Sample Output 2:
404>

题意:

给出l位数, 求最近的(从左到右)k位素数, 如果没有输出404
细节: 考虑0023这样的情况

题解:

水题…但还是始终有一道测试点没过去, 而且还是要熟悉一下substr和stoi的应用
有一个测试点一直过不去…后来发现对于0023这样的情况要输出整个字符串, 而不是int

#include <iostream>
#include <string>
using namespace std;
bool isPrime(int n){
    if(n == 0 || n == 1) return false;
    for(int i = 2; i * i <= n; i++)
        if(n % i == 0) return false;
    return true;
}
int main()
{
    int l, k;
    string s;
    cin >> l >> k >> s;
    for(int i = 0; i <= l-k; i++){
        string t = s.substr(i, k);
        int num = stoi(t);
        if(isPrime(num)){
            cout << t; //输出字符串而不是输出数 !
            return 0;
        }
    }
    cout << "404\n";

	return 0;
}

PAT 1152 Google Recruitmen (水题 细心)