PAT A1152 Google Recruitment (20point(s))

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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.

题意

给出一个l长度的字符串,求出第一个k位的素数(最高位开始)

解答

从最高位开始,枚举长度为k字串(使用string类的substr函数),并转化为整数(可以使用stoi函数也可以自己编写一个函数)

#include<iostream>
#include<string>
#include<cmath>

using namespace std;

long long strToInteger(string s, int begin, int end)
{
	long long res = 0;
	for (int i = begin; i < end; i++) {
		res = res * 10 + s[i] - '0';
	}
	return res;
}
bool isPrime1(long long d)
{
	long long end = (int)sqrt(d);
	for (int i = 2; i <= end; i++)
		if (d % i == 0) return false;
	return true;
}

bool isPrime2(long long d)
{
	if (d == 2 || d == 3) return true;
	if (d % 6 != 1 && d % 6 != 5) return false;
	long long end = (int)sqrt(d);
	for (int i = 5; i <= end; i += 6)
		if (d % i == 0 || d % (i + 2) == 0) return false;
	return true;
}
int main()
{
	string str, head;
	int l, k;
	cin >> l >> k >> str;
	for (int i = 0; i <= l - k; i++) {
		long long d = strToInteger(str, i, i + k);//stoi(str.substr(i, k));
		if (isPrime2(d)) {
			cout << str.substr(i, k);
			return 0;
		}
	}
	cout << "404";
	return 0;
}

prime2函数是prime1函数的优化版本,该函数能够更快地判断一个整数是否为质数,证明如下:
如果一个整数能够表示为6x+2(被2整除)、6x+3(被3整除)、6x+4(被2整除),那这个数一定不是质数,因此质数只可能存在于6x-1 和 6x+1(5,7, 11,13...)和(2,3)中,在for循环中就可以将步长设为6,以此来加快运行速度。