HDU 4861（多校）1001 Couple doubi

最新推荐文章于 2019-10-16 08:31:42 发布

转载最新推荐文章于 2019-10-16 08:31:42 发布 · 78 阅读

本文介绍了一个简单的游戏胜负判断算法，游戏中玩家轮流从桌上取球，每个球有一个基于质数p的特定值，最终比较双方得分确定胜者。

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
   

Source

2014 Multi-University Training Contest 1

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没听过什么费马定理，就仅仅知道这：

卧槽，就这样勉强写的，哎，说多了都是泪。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

long long  k,p;
int main()
{
    while(~scanf("%I64d%I64d",&k,&p))
    {
        int s=0;
        if(p==2)
        {
           if(((k+1)/2%2))
              cout<<"YES"<<endl;
           else
              cout<<"NO"<<endl;
           continue;
        }
        s=k/(p-1);
        if(s%2)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

事实上仅仅要这样。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

long long  k,p;
int main()
{
    while(~scanf("%I64d%I64d",&k,&p))
    {
        int s=k/(p-1);
        if(s%2)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

我小学没毕业，干只是那些高中生。。