1015. Reversible Primes (20)

最新推荐文章于 2024-02-22 13:10:55 发布

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最新推荐文章于 2024-02-22 13:10:55 发布

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原文链接：http://www.cnblogs.com/yuxiaoba/p/8554872.html

本文介绍了一种算法，用于判断一个正整数N在给定的数制D下是否为可逆素数。即该数及其在指定进制下的反转形式均为素数。通过分解整数、反转并重新计算其在不同数制下的值来实现。

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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:

Sample Output:

Yes
Yes
No

题目大意：判断原数字是否是素数，倒序后是否是素数，如果都是素数输出Yse

这里的是指原数字与它在数制下的倒序，如23 2，是10111（2）的倒叙11101的真实数字。

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<stdlib.h>
 4 int in[6];
 5 int main()
 6 {
 7     int n,d,m,temp;
 8     int i,j;
 9     int sum,flag;
10     while( scanf("%d",&n)!=EOF)
11     {
12         if( n<0 ) break;
13         scanf("%d",&d);
14         i=0;  //分解整数的下标
15         sum=0;  //转换后的值
16         flag = 0;  //标记是否能转换
17         temp=n;   //保存n的值
18         
19         while( n ) //分解位数
20         {
21             in[i++]=n%d;
22             n /= d;
23         }
24         for( j=0; j<i ; j++) //求转换的值
25             sum = sum*d+ in[j];
26         if( isPrime(sum) && isPrime(temp)) //如果转换前和转换后都是素数
27             flag =1;
28         if( !flag) printf("No\n");
29         else printf("Yes\n");
30     }
31     return 0;
32 }
33 
34 int isPrime( int sum)
35 {
36     int i,m;
37     if( sum<2 ) return 0;
38     if( sum==2 || sum ==3 )return 1;
39     m =(int) sqrt(sum)+1;
40     for( i=2; i<m; i++)
41         if( sum%i==0 )
42             return 0;
43     return 1;
44 }