【hdu 3951】Coin Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1851 Accepted Submission(s): 1065
Problem Description
After hh has learned how to play Nim game, he begins to try another coin game which seems much easier.
The game goes like this:
Two players start the game with a circle of n coins.
They take coins from the circle in turn and every time they could take 1~K continuous coins.
(imagining that ten coins numbered from 1 to 10 and K equal to 3, since 1 and 10 are continuous, you could take away the continuous 10 , 1 , 2 , but if 2 was taken away, you couldn’t take 1, 3, 4, because 1 and 3 aren’t continuous)
The player who takes the last coin wins the game.
Suppose that those two players always take the best moves and never make mistakes.
Your job is to find out who will definitely win the game.
Input
The first line is a number T(1<=T<=100), represents the number of case. The next T blocks follow each indicates a case.
Each case contains two integers N(3<=N<=109,1<=K<=10).
Output
For each case, output the number of case and the winner “first” or “second”.(as shown in the sample output)
Sample Input
2
3 1
3 2
Sample Output
Case 1: first
Case 2: second
【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=3951
【题解】
会发现如果k!=1
先手不管怎样选一段;
我们总能在这n个硬币所围成的对角线上选取同样的一段(或者比它多一个或少一个->n为奇数的情况);这样就分成了两段硬币;
这两段硬币的个数相同;
->如果对应两个游戏的话,对应的sg函数必然是相同的;
则sg[x]^sg[x]==0
所以先手输;
我们总能让第一个人面对这种情况;
所以第一个人总是输的;
但是k=1的时候就不一定可以;
只有当n为偶数的时候第一个人输;n为奇数的话第一个人无论如何都会赢的.
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int T;
int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(T);
rep1(ii,1,T)
{
int n,k;
rei(n);rei(k);
printf("Case %d: ",ii);
if (k>=n || (k==1 && n&1))
puts("first");
else
puts("second");
}
return 0;
}
相关文章
- HDU 1556-差分数组和线段树的对比分析-Color the ball
- hdu 3980 Paint Chain(SG函数)
- HDU 3336 KMP算法中对next数组的理解「建议收藏」
- HDU P3341 Lost’s revenge 题解+数据生成器
- hdu 3336 Count the string(kmp应用)
- hdu 3336 Count the string 用心写的题解
- hdu 1142_hdu1001
- HDU 3336 Count the string 解题报告
- [ACM] HDU 1006 解题报告
- H - Partial Tree HDU - 5534 【 完全背包 】
- HDU-2084 数塔
- 敌兵布阵(HDU 1166)
- I Hate It (HDU 1754)
- 校庆神秘建筑(HDU 1411)
- 还是畅通工程(HDU 1233)
- Friend-Graph (HDU 6152)2017中国大学生程序设计竞赛 - 网络选拔赛
- CaoHaha's staff (HDU 6154)(2017中国大学生程序设计竞赛 - 网络选拔赛)
- Oil Deposits (HDU - 1241 )(DFS思路 或者 BFS思路)
- 畅通工程续(HDU 1874)(简单最短路)
- Drainage Ditches (HDU - 1532)(最大流)
- Ancient Go ( HDU - 5546 ) ( BFS 搜索是否相连)
- 统计难题 【 HDU - 1251 】【 字典树 】
- Phone List 【 HDU - 1671 】 【 字典树判断是否存在前缀 】
- A * B Problem Plus【HDU 1402】 【FFT求高精度乘法】
- Bone Collector HDU - 2602【 01 背包 】