HDU-4861 Couple doubi

http://acm.hdu.edu.cn/showproblem.php?pid=4861

                            Couple doubi

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 675    Accepted Submission(s): 484

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

分析：找规律题，以p-1为一个周期，在p-1的这个位置是数字，其余为0，所以是奇数，就输出yes，否则就no。

这种题一定耐心的把所有情况写出来，看之间的规律。

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

}