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”.
(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).
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
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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;
}