PAT甲级 1152 Google Recruitment (20 分)

最新推荐文章于 2022-02-11 10:27:53 发布

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最新推荐文章于 2022-02-11 10:27:53 发布

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本文链接：https://blog.youkuaiyun.com/DoMoreSpeakLess/article/details/119518650

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该博客介绍了一个编程挑战，即从一个给定的数字中找出连续的K位素数。文章提供了一个C++代码示例，用于解决这个问题，并展示了如何通过遍历和素数检查函数来找到符合条件的素数。在示例输入中，对于205位数字23654987725541023819，找到了第一个23位的素数49877。如果无法找到满足条件的素数，则输出404。

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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.

Sample Input 1:

20 5
23654987725541023819

Sample Output 1:

Sample Input 2:

10 3
2468024680

Sample Output 2:

#include <cstdio>
#include <cmath>

using namespace std;
char digit[1005];
int n, m;

bool isPrime(long long n) {
    if (n <= 1) return false;
    long long sqr = (long long) sqrt(1.0 * n);
    for (int i = 2; i <= sqr; ++i) {
        if (n % i == 0)return false;
    }
    return true;
}

long long int toNum(int b, int e) {
    if (e > n) return 1;
    int num = 0;
    for (int i = b; i < e; ++i) {
        num = num * 10 + (digit[i] - '0');
    }
    return num;
}

int main() {
    scanf("%d %d", &n, &m);
    scanf("%s", digit);
    for (int i = 0; i < n; ++i) {
        if (isPrime(toNum(i, i + m))) {
            for (int j = 0; j < m; ++j) {
                printf("%c", digit[i + j]);
            }
            return 0;
        }
    }
    printf("404");
    return 0;
}