2014 Multi-University Training Contest 1/HDU4861_Couple doubi(数论/法)
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

找规律找到的,,,sad,,,
非常easy找到循环节为p-1,每个循环节中有一个非零的球。所以仅仅要推断有多少完整循环节。在推断奇偶,。,
#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
(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”.
In the first line there are two Integers k and p(1<k,p<2^31).
2 3
20 3
YES
NO
2014 Multi-University Training Contest 1/HDU4861_Couple doubi(数论/法)的更多相关文章
- 2019 Multi-University Training Contest 1 - 1011 - Function - 数论
http://acm.hdu.edu.cn/showproblem.php?pid=6588 新学到了一个求n以内与m的gcd的和的快速求法.也就是下面的S1. ①求: $ \sum\limits_{ ...
- HDU4888 Redraw Beautiful Drawings(2014 Multi-University Training Contest 3)
Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Jav ...
- hdu 4946 2014 Multi-University Training Contest 8
Area of Mushroom Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
- 2014 Multi-University Training Contest 9#11
2014 Multi-University Training Contest 9#11 Killing MonstersTime Limit: 2000/1000 MS (Java/Others) ...
- 2014 Multi-University Training Contest 9#6
2014 Multi-University Training Contest 9#6 Fast Matrix CalculationTime Limit: 2000/1000 MS (Java/Oth ...
- 2014 Multi-University Training Contest 1/HDU4864_Task(贪心)
解题报告 题意,有n个机器.m个任务. 每一个机器至多能完毕一个任务.对于每一个机器,有一个最大执行时间Ti和等级Li,对于每一个任务,也有一个执行时间Tj和等级Lj.仅仅有当Ti>=Tj且Li ...
- hdu 4937 2014 Multi-University Training Contest 7 1003
Lucky Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) T ...
- hdu 4941 2014 Multi-University Training Contest 7 1007
Magical Forest Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Other ...
- hdu 4939 2014 Multi-University Training Contest 7 1005
Stupid Tower Defense Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/ ...
随机推荐
- tb连续aaaaa123aaa自适应
在连续的字符串数字中,td不会自适应大小,需要加上样式 style="word-break : break-all; overflow:hidden; " <table> ...
- iOS8推送消息的回复处理速度
iOS8我们有一个新的通知中心,我们有一个新的通报机制.当在屏幕的顶部仅需要接收一个推拉向下,你可以看到高速接口,天赋并不需要输入应用程序的操作.锁定屏幕,用于高速处理可以推动项目. 推送信息,再次提 ...
- 开源 自由 java CMS - FreeCMS1.9 分纪录
项目地址:http://www.freeteam.cn/ 2.4.1 积分记录 查看系统全部会员积分记录. 从左側管理菜单点击积分记录进入. 版权声明:本文博客原创文章,博客,未经同意,不得转载.
- WPF中两条路径渐变的探讨
原文:WPF中两条路径渐变的探讨 我们在WPF中,偶尔也会涉及到两条路径作一些“路径渐变 ”.先看看比较简单的情形:如下图(关键点用红色圆点加以标识):(图1) 上面图1中的第1幅图可以说是最简单的路 ...
- Xamarin.Android 入门实例(4)之实现对 SQLLite 进行添加/修改/删除/查询操作
1.Main.axml <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns: ...
- tolower (Function)
this is a function that Convert uppercase letter to lowercase Converts c to its lowercase equivalent ...
- CSS设计指南之定位
原文:CSS设计指南之定位 CSS布局的核心是position属性,对元素盒子应用这个属性,可以相对于它在常规文档流中的位置重新定位.position属性有4个值:static.relative.ab ...
- CCLayer在Touch事件(Standard Touch Delegate和Targeted Touch Delegate)
在做练习,触摸故障,看到源代码,以了解下触摸事件. 练习操作:直CClayer子类init在 this->setTouchEnabled(true); 事件处理方法覆盖 virtual bool ...
- 连接字符串中Min Pool Size的理解是错误,超时时间已到,但是尚未从池中获取连接。出现这种情况可能是因为所有池连接均在使用,并且达到了最大池大小。
Min Pool Size的理解是错误的 假设我们在一个ASP.NET应用程序的连接字符串中将Min Pool Size设置为30: <add name="cnblogs" ...
- 有向图的邻接矩阵表示法(创建,DFS,BFS)
package shiyan; import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; publi ...