- 题意:给定数10进制数N和进制D,将N转化成D进制,再将D进制的数反转,最后转化为10进制数M,判断N和M是否都是质数。
#include <iostream>
using namespace std;
int isPrime(int n) {
if(n < 2)
return 0;
for(int i = 2; i*i <= n; ++i) {
if(n%i == 0)
return 0;
}
return 1;
}
int change(int n, int d) {
int b[100], count = 0;
while(n) {
b[count] = n%d;
count++;
n /= d;
} // 得到的b[]正序就是转换后的数的逆序
int decimal = 0;
for(int i = count-1; i >= 0; --i) {
int temp = 1;
for(int j = 0; j < count-1-i; ++j) {
temp = temp * d;
}
decimal = decimal + b[i] * temp;
}
return decimal;
}
int main(int argc, const char * argv[]) {
int N, D;
while(cin >> N) {
if(N < 0)
break;
cin >> D;
int re = change(N, D);
if(isPrime(N) && isPrime(re))
cout << "Yes"<<endl;
else
cout << "No"<<endl;
}
return 0;
}
附原题:
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