1015 Reversible Primes (20)(20 分)
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 (< 10^5) 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
思路:
首先用素数筛筛出素数,然后原数判断一次,利用栈把原数倒一下,倒过来的数再判断一次。
代码:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define MAX 100000
int su[MAX], cnt;
bool isprime[MAX];
stack<int> s;
void prime()
{
memset(isprime, true, sizeof(isprime));
isprime[0] = isprime[1] = false;
cnt = 1;
for (int i = 2; i<MAX; i++)
{
if (isprime[i])
su[cnt++] = i;
for (int j = 1; j < cnt&&su[j] * i < MAX; j++)
isprime[su[j] * i] = false;
}
}
int main()
{
prime();
int n, d;
while (cin >> n && n >= 0)
{
cin >> d;
if (isprime[n])
{
while (n)
{
s.push(n%d);
n /= d;
}
int mul = 1;
while (!s.empty())
{
n += mul * s.top();
s.pop();
mul *= d;
}
if (isprime[n])
{
cout << "Yes" << endl;
continue;
}
}
cout << "No" << endl;
}
return 0;
}