merge branches

PATAdvanced/1015.c

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+/**
+ * 1015. Reversible Primes (20)
+ * 
+ * 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
+ */
+
+#include <stdio.h>
+
+int iPrime(int N)
+{
+    if(N == 0 || N == 1)
+        return 0;
+    for(int i = 2; i * i <= N; i++)
+        if(N % i == 0)
+            return 0;
+    return 1;
+}
+
+int Rev(int N, int D)
+{
+    int Nrev;
+    for(Nrev = 0; N; N /= D)
+    {    
+        Nrev *= D; 
+        Nrev += N % D;
+    }
+    return Nrev;
+}
+
+int main()
+{
+    int N, D;
+    scanf("%d", &N);
+    while(N >= 0)
+    {
+        scanf("%d", &D);
+        puts(iPrime(N) && iPrime(Rev(N, D)) ? "Yes" : "No");
+        scanf("%d", &N);
+    }
+    
+    return 0;
+}

README.md

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 欢迎讨论 oliver_lew@outlook.com
 
-(PAT B已经刷完1001-1080，PAT A刚刚开始)
-
 * 简书中写了代码的解释
-  * [PAT Basic](https://www.patest.cn/contests/pat-b-practise): http://www.jianshu.com/p/c2b557516b50
-  * [PAT Advanced](https://www.patest.cn/contests/pat-a-practise): http://www.jianshu.com/p/8944b15f8194
+  * [PAT Basic](https://www.patest.cn/contests/pat-b-practise) (已经刷完1001-1080): http://www.jianshu.com/p/c2b557516b50
+  * [PAT Advanced](https://www.patest.cn/contests/pat-a-practise) (刚刚开始): http://www.jianshu.com/p/8944b15f8194