题干前半略。

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

#include<stdio.h>
#include<vector>
#include<string>
#include<iostream>
#include<math.h>
using namespace std; bool isprime(int n)
{ if(n==1||n==0) return false;
for(int i=2;i<=(int)sqrt(n);i++)
{
if(n%i==0) return false;
}
return true;
}
bool flag;
int main()
{
int num;
int primenum;
scanf("%d",&num);
scanf("%d",&primenum);
string seq;
cin>>seq;
while(seq[0]=='0') seq.erase(seq.begin());
int i;
for(i=0;i<num;i++)
{
int start=i;
char temp[20];
int t=0;
if(num-i<primenum) break;
for(int j=i;j<i+primenum;j++)
{
temp[t++]=seq[j];
}
int newnum;
sscanf(temp,"%d",&newnum);
temp[t]='\0';
if(isprime(newnum))
{
cout<<temp;
flag=true;
return 0;
}
}
if(i==num-primenum+1) printf("404\n");
}

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