Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 221 Accepted Submission(s): 186
DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on the desk. Every ball has a value and the value of ith (i=1,2,…,k) ball is 1^i+2^i+…+(p-1)^i (mod p). Number p is a prime number that is chosen by DouBiXp and his girlfriend. And then they take balls in turn and DouBiNan first. After all the balls are token, they compare the sum of values with the other ,and the person who get larger sum will win the game. You should print “YES” if DouBiNan will win the game. Otherwise you should print “NO”.
Multiply Test Cases. In the first line there are two Integers k and p(1<k,p<2^31).
For each line, output an integer, as described above.
2 3 20 3
1^k + 2^k + 3^k + (p-1)^k % p 只有当k|p-1的时候才等于p-1,其余都是0的。 故非零的个数是 k/(p-1) 判断这个是奇数还是偶数就可以了。