Couple doubi（费马小定理或打表找规律+简单博弈）

Link：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): 1434    Accepted Submission(s): 993

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
  

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

/*附上打表找规律代码： 
#include<stdio.h>
#include<math.h>
int main ()
{
   int i ,j ,sum;
   printf("9 3\n");
   for(i = 1 ;i <= 9 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 3 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3;
      printf("%d " ,sum % 3);
   }
   printf("\n2 3\n");
   for(i = 1 ;i <= 2 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 3 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3;
      printf("%d " ,sum % 3);
   }
   printf("\n9 2\n");
   for(i = 1 ;i <= 9 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 2 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 2;
      printf("%d " ,sum % 2);
   }
   printf("\n10 3\n");
   for(i = 1 ;i <= 10 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 3 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3;
      printf("%d " ,sum % 3);
   }
   
   printf("\n10 5\n");
   for(i = 1 ;i <= 10 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 5 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 5;
      printf("%d " ,sum % 5);
   }
   
   printf("\n10 7\n");
   for(i = 1 ;i <= 10 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 7 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 7;
      printf("%d " ,sum % 7);
   }
   
    printf("\n10 11\n");
   for(i = 1 ;i <= 10 ;i ++)
   {
      sum = 0;
      for(j = 1 ;j < 11 ;j ++)
      sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 11;
      printf("%d " ,sum % 11);
   }
   
   getchar();
}*/

/*编程思想：多试几个很容易可以发现规律。。。只有在p-1的倍数时有非0元素存在。
那么由于是博弈，大家每次肯定都是选那个比较大的数，所以，若是 k/(p-1)为奇数则先手必胜，否则平局*/
//AC CODE: 
#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;
}