1152 Google Recruitment （20 分）前导0，快速判单个素数_pow(10, k

本文探讨了一道谷歌发布的招聘谜题，挑战者需在给定的长数字串中找出第一个K位的素数。文章介绍了如何通过高效算法筛选素数，并提供了解题的C++代码实现，包括快速判断素数的方法和处理前导零的技巧。

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前导0, 0023不能23
素数更快判断，利用素数>3后总在6左右，如5,7详解
补齐位数是代数时，用printf("%0*d",k,n);
低级错误，pow(10,k-1)不是k-1。。

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.

在这里插入图片描述

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

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int l,k;
char t;
int fig=0;
bool prime(int n)//快速判断
{
	int n_sqrt;
	if(n==2 || n==3) return 1;
	if(n==1||n%6!=1 && n%6!=5) return 0;
	n_sqrt=sqrt((float)n);
	for(int i=5;i<=n_sqrt;i+=6)
	{
	    if(n%(i)==0 || n%(i+2)==0) return 0;
	}
        return 1;
}
int main()
{
	scanf("%d%d%c",&l,&k,&t);
	int Mod=pow(10,k-1);
	for(int i=1;i<=l;i++)
	{
		scanf("%c",&t);
		fig=fig*10+(t-'0');
		if(i>=k)
		{
			if(prime(fig))
			{
			//	while(fig/Mod==0)//前导0 
			//	{
			//		printf("0");
			//		Mod/=10;
			//	}//
				printf("%0*d\n",k,fig);//直接输出前导0 
				return 0;
			}
			fig=fig%Mod;
		}
	}
	printf("404\n");
	return 0;
}