Couple doubi（打表找规律，数论）

Problem Description
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”.
 

Input
Multiply Test Cases. 
In the first line there are two Integers k and p(1<k,p<2^31).
 

Output
For each line, output an integer, as described above.
 

Sample Input
2 3

20 3
 

Sample Output
YES

NO

打表找规律真没看出来（反思）

#include <iostream>
#include <cstdio>
using namespace std;
 
int main()
{
    int k, p;
    while(~scanf("%d%d", &k, &p))
    {
        if((k/(p-1))%2==1)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}