PAT甲级练习1015. Reversible Primes (20)

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

主要分为两部分，一部分是素数的判断，一部分是进制间的转化，总体来说还是比较简单的，但当中还是有些小问题

#include <cstdio>#include <cmath>#include <algorithm>#include <vector>using namespace std;int n, d;vector<int> num;bool isPrime(int x){if(x<=1)return false;for(int i=2; i<=int(sqrt(1.0*x)); i++){//这里我一开始 i<=sqrt(1.0*x)+1 通不过测试点1，是因为类型不匹配吗？if(x%i==0) return false;}return true;}int trans(int x, int y){int i=0;do{num.push_back(x%y);x /= y;}while(x!=0);for(int j=0; j<num.size(); j++){i = i * y + num[j];printf("%d", i);}return i;}int main(){scanf("%d", &n);while(n>0){vector<int> blank;num.swap(blank);scanf("%d", &d);if(isPrime(n) && isPrime(trans(n, d)))printf("Yes\n");elseprintf("No\n");scanf("%d", &n);}return 0;}