一、题目描述
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
二、解题思路
对于这道题，我们可以写两个函数，一个用来把字符串转换成数字，一个用来判断数字是否为素数，接着对于输入的长字符串，我们进行遍历，每一次取一段字符子串，转化成数字，然后判断是否为素数，如果找到了，就直接输出然后return，如果没有找到，在最后输出404即可。
三、AC代码
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
LL toNum(string tmp) //将字符串转化成数字的函数
{
int sze = tmp.size();
LL ans = 0;
for(int i=0; i<sze; i++)
{
ans = ans*10 + tmp[i] - '0';
}
return ans;
}
bool checkprime(LL num) //判断素数
{
for(int i=2; i*i <= num; i++)
{
if(num %i == 0) return false;
}
return true;
}
int main()
{
int n, K;
string str;
scanf("%d%d", &n, &K);
cin >> str;
for(int i=0; i+K <= n; i++)
{
string tmp = str.substr(i, K); //子字符串
LL num = toNum(tmp);
if(checkprime(num))
{
cout << tmp << endl;
return 0;
}
}
printf("404\n");
return 0;
}