pat 1152

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

 

 

题意：求一个给定字符串中最先的连续k位的素数，如果不存在返回404

思路：字符数组存储字符串，设置一个指向第一个字符的变量，依次检测从该地方起的k位是否是素数。

注意点：有些素数以0开头，输出的时候前面的0不能省略，而素数的位数根据k来确定，但k又是输入进来的，因此这里需要特殊处理下

代码如下

#include<cstdio>
#include<math.h>
char num[1005];
bool isPrime(int sum){
    int mark=0;
    for(int i=2;i<sqrt(sum);i++){
        if(sum%i==0){
            mark=1;
            break;
        } 
    }
    if(mark==1)
        return false;
    else
        return true; 
}
long long fun(int i,int k){
    long long sum=0;
    for(int j=i;j<k+i;j++){
        sum=sum*10+num[j]-'0'; 
    }
    if(isPrime(sum))
        return sum;
    else
        return -999999l;
}
int main(){
    int l,k;
    long long ans;
    scanf("%d%d",&l,&k);
    scanf("%s",num);
    int mark=0;
    for(int i=0;i<l-k+1;i++){
        if(fun(i,k)!=-999999l){
            mark=1;
            
            ans=fun(i,k);
            break;
        }
    }
    if(mark==0)
        printf("404");
    else{
        int size=0;
        long long temp=ans;
        while(temp!=0){
            temp/=10;
            size++;
        }
        if(size!=k){
            for(int i=0;i<k-size;i++){
                printf("0");
            }
        }
        printf("%d",ans);
    }
    return 0;
}