HDU4861 Couple doubi

最新推荐文章于 2016-02-07 20:41:13 发布

原创最新推荐文章于 2016-02-07 20:41:13 发布 · 561 阅读

0 ·

CC 4.0 BY-SA版权

2014多校同时被 2 个专栏收录

17 篇文章

订阅专栏

数论

4 篇文章

订阅专栏

本文探讨了游戏开发中涉及的策略与算法实现，包括数值分析、数据结构、算法优化等核心内容，深入剖析游戏开发过程中的关键技术和实战案例。

Couple doubi

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 428 Accepted Submission(s): 328

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

题意：有k个球，两个人去取，每个球有个价值，谁取到最后得到的价值最大就赢，每个就得价值有个公式算。

分析：官方题解有正确解法，当初是写了个暴力程序，然后看出了规律，就做出来了

#include<cstdio>
using namespace std;
int main()
{
    int n,p;
    while(scanf("%d%d",&n,&p)==2)
    {
        n=n/(p-1);
        if(n&1)
        {
            printf("YES\n");
        }
        else
            printf("NO\n");
    }
    return 0;
}