hdu4861 Couple doubi---2014 Multi-University Training Contest 1

题目链接：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): 238    Accepted Submission(s): 199

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

 

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严格来讲，这应该是道数学题。但是假设是推导的话，数学沫沫表示死都推不出来。。。

于是果断。打几个小数据找规律了。

。。

。

20 3
0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2
20 5
0 0 0 4 0 0 0 4 0 0 0 4 0 0 0 4 0 0 0 4

多试几个非常easy能够发现规律。。。仅仅有在p-1的倍数时有非0元素存在。

那么因为是博弈，大家每次肯定都是选那个比較大的数，所以。若是 k/(p-1)为奇数则先手必胜。否则平局

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<bitset>
using namespace std;
int main(){
    int k,p;
    while(cin>>k>>p){
        if(k/(p-1)&1) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}