Gerg's Cake

Gerg is having a party, and he has invited his friends. p of them have arrived already, but a are running
late. To occupy his guests, he tried playing some team games with them, but he found that it was
impossible to divide the p guests into any number of equal-sized groups of more than one person.
Luckily, he has a backup plan | a cake that he would like to share between his friends. The cake is
in the shape of a square, and Gerg insists on cutting it up into equal-sized square pieces. He wants to
reserve one slice for each of the a missing friends, and the rest of the slices have to be divided evenly
between the p remaining guests. He does not want any cake himself. Can he do it?
Input
The input will consist of several test cases. Each test case will be given as a non-negative integer a and
a positive integer p as specied above, on a line. Both a and p will t into a 32-bit signed integer. The
last line will contain `-1 -1' and should not be processed.
Output
For each test case, output `Yes' if the cake can be fairly divided and `No' otherwise.
Sample Input
1 3
1024 17
2 101
0 1
-1 -1
Sample Output
Yes
Yes
No
Yes

题意：输入a,p,问把一个正方形蛋糕分成若干个小正方形，然后分给还没到的a个人一人一个后再使得已经到的p个人正好平分。

分析：本题容易被样例误导，以为只要a+p是素数就输出“Yes”。实际上并不是这么一个回事。

假设切完之后的蛋糕有x*x个，那么就有x*x=a+n*p，其中n是一个整数。

我们对这个式子左右两边同时求余就有：x*x = a (mod p)。

联系费马小定理：x^(p-1) =1 (mod p).

就有：x^(p-1)=a^(p-1)/2=1　  (mod p)

现在只要计算：a^(p-1)/2 =1 (mod p)是否成立即可。

实现过程写一个快速幂就行了，不过要注意的是底数a进行求幂运算之前应该先进行a%p否则有可能会因为溢出导致WA

 

AC code: