PAT -- 甲级1015（1015 Reversible Primes）
1015 Reversible Primes (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 NNN (<105< 10^5<105) and DDD (1<D≤101 < D \le 101<D≤10), you are supposed to tell if NNN is a reversible prime with radix DDD.
Input Specification:
The input file consists of several test cases. Each case occupies a line which contains two integers NNN and DDD. The input is finished by a negative NNN.
Output Specification:
For each test case, print in one line Yes if NNN is a reversible prime with radix DDD, or No if not.
Sample Input:
73 10
23 2
23 10
-2
Sample Output:
Yes
Yes
No
给两个数n，radix分别表示一个十进制数以及一个基数，判断一下该数转化为radix进制并反转后再化为十进制数，判断这两个数是否都为素数。Yes或No，当n为负数的时候结束输入。注意0与1输出No
思路：数据量在1e5以内，并且基数不会超过10，所以转化的数也不会超过maxn。然后 O(1)判断。
Code：
#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)1e5+5;
bool vis[maxn];
int prime[maxn], len;
void euler(int n) {
memset(vis, false, sizeof(vis));
vis[0] = vis[1] = true;
int k, i, j;
len = 0;
for (i = 2; i <= n; i++) {
if (!vis[i]) prime[++len] = i;
for (j = 1; j <= len; j++) {
k = i * prime[j];
if (k > n) break;
vis[k] = true;
if (i % prime[j] == 0) break;
}
}
}
//将n转为m进制的反转数
int change(int n, int m) {
int ans = 0;
while (n > 0) {
ans = ans * m + n % m;
n /= m;
}
return ans;
}
int main() {
int data,radix;
euler(maxn-1);
while (scanf("%d", &data) != EOF && data >= 0) {
scanf("%d", &radix);
if (!vis[data] && !vis[change(data, radix)]) {
printf("Yes\n");
} else {
printf("No\n");
}
}
return 0;
}
