本文主要是介绍2014 Multi-University Training Contest 1/HDU4861_Couple doubi(数论/规律)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
解题报告 两人轮流取球，大的人赢，，， 贴官方题解，，，反正我看不懂，，，先留着理解 关于费马小定理 关于原根 
找规律找到的，，，sad，，， 很容易找到循环节为p-1，每一个循环节中有一个非零的球，所以只要判断有多少完整循环节，在判断奇偶，，， #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>using namespace std;int main()
{int k,p,i,j;while(scanf("%d%d",&k,&p)!=EOF){int q=k/(p-1);if(q%2==0)printf("NO\n");else printf("YES\n");}return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>using namespace std;int main()
{int k,p,i,j;while(scanf("%d%d",&k,&p)!=EOF){int q=k/(p-1);if(q%2==0)printf("NO\n");else printf("YES\n");}return 0;
}
Couple doubi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 145 Accepted Submission(s): 122
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).
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
Source
2014 Multi-University Training Contest 1
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