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 1023 223 10-2

Sample Output:

YesYesNo

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提交代码

题意:给你一个数n,n是一个10进制数,问n是否为素数且n在d进制下反转过来(比如说13二进制形式为1101, 反转过来是1011就变成了11)是否是素数.

题解:这是一道模拟题, 先用个数组存下d进制下的数,然后在反转,在转变成10进制,看其是否为素数就好了

#include <cstdio>bool is_prime(int n) {    if (n < 2) return false;    for (int i = 2; i * i <= n; ++i)        if (n % i == 0)            return false;    return true;}bool check(int n,int d) {    int cnt = 0;    int ary[50];    int tmp1 = n;    while (tmp1) {        ary[cnt++] = tmp1 % d;        tmp1 /= d;    }    int tmp2 = 0;    int x = 1;    for (--cnt; ~cnt; --cnt, x *= d)        tmp2 += x * ary[cnt];    return is_prime(n) && is_prime(tmp2);}int main() {    int n, d;    while (~scanf("%d%d", &n, &d), n >= 0) {        if (check(n, d) == true)            puts("Yes");        else            puts("No");    }    return 0;}