PAT (Advanced Level) Practise 1015. Reversible Primes (20) 素数筛法 进制转换

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 10⁵) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line "Yes" if N is a reversible prime with radix D, or "No" if not.

Sample Input:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

素数筛法判定素数，然后用一个数组存反转的数，然后进制转换再判定一次素数即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 100000;
bool notp[maxn + 5];

void getP() {
	memset(notp, false, sizeof(notp));
	notp[0] = true;
	notp[1] = true;
	for (int i = 2; i <= maxn; i++) {
		if (!notp[i]) {
			for (int j = 2; i * j <= maxn; j++) {
				notp[i * j] = true;
			}
		}
	}
}

bool isRe(int N, int D) {
	int re[100], len, ans;
	len = 0;
	do {
		re[len++] = N % D;
		N /= D;
	} while (N);
	ans = 0;
	for (int i = 0; i < len; i++) {
		ans *= D;
		ans += re[i];
	}
	return !notp[ans];
}

int main()
{
	int N, D;
	getP();
	while (1) {
		scanf("%d", &N);
		if (N < 0) break;
		scanf("%d", &D);
		if (notp[N] || !isRe(N, D)) {
			puts("No");
		}
		else {
			puts("Yes");
		}
	}
	return 0;
}