zoj 2507 Let's play a game
2015-08-21 19:40
393 查看
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1507
Elves from the Lothvain forest have created a very interesting game. The rules are very simple:
There are two players.
In the beginning there are n groups of stones. The i-th group contains a_i stones.
A player has to take a positive number of stones from exactly one of the groups in his turn.
A player who can’t make a move (each group is empty) wins.
Very soon they have learnt how to always make the best possible move. Your task is to write a program that computes which of the players has a winning strategy for a given situation.
INPUT
The first number appearing in the input is number of datasets t (5<=t<=50). Each dataset is given in following format. The first row contains a single number n (1<=n<=10000). The next row contains exactly n numbers: a_1, a_2,…, a_n (0<=a_i<=1000000000). There are no empty lines between the datasets. You may assume that at least one group of stones is not empty.
OUTPUT
For each of datasets your program should write:
1 if the first player has a winning strategy
2 otherwise
in a separate row.
SAMPLE INPUT
2
3
1 1 1
5
1 0 1 1 1
SAMPLE OUTPUT
2
1
题意:NIM游戏的变种,胜利状态变为不能取的取胜。
先手必胜当且仅当:
1、若所有堆的石子最大数量为1,且数量为1的堆数为偶数
2、若有的堆数量超过1,SG为非0
证明:情况一、很容易想到。
情况二、先分析只有一堆石子的数量大于1,那么这个人可以控制后手的1的数量,因此是必胜态(SG必为非0)。
其次,如果有大于1堆的石子的数量大于2,那么如果SG为非0,那么先手只要将SG变为0,后手无论进行什么操作,SG都会变成非0(异或的性质),这样进行下去,终会达到必胜态(只有一堆石子的最大数量大于1)。
Elves from the Lothvain forest have created a very interesting game. The rules are very simple:
There are two players.
In the beginning there are n groups of stones. The i-th group contains a_i stones.
A player has to take a positive number of stones from exactly one of the groups in his turn.
A player who can’t make a move (each group is empty) wins.
Very soon they have learnt how to always make the best possible move. Your task is to write a program that computes which of the players has a winning strategy for a given situation.
INPUT
The first number appearing in the input is number of datasets t (5<=t<=50). Each dataset is given in following format. The first row contains a single number n (1<=n<=10000). The next row contains exactly n numbers: a_1, a_2,…, a_n (0<=a_i<=1000000000). There are no empty lines between the datasets. You may assume that at least one group of stones is not empty.
OUTPUT
For each of datasets your program should write:
1 if the first player has a winning strategy
2 otherwise
in a separate row.
SAMPLE INPUT
2
3
1 1 1
5
1 0 1 1 1
SAMPLE OUTPUT
2
1
题意:NIM游戏的变种,胜利状态变为不能取的取胜。
先手必胜当且仅当:
1、若所有堆的石子最大数量为1,且数量为1的堆数为偶数
2、若有的堆数量超过1,SG为非0
证明:情况一、很容易想到。
情况二、先分析只有一堆石子的数量大于1,那么这个人可以控制后手的1的数量,因此是必胜态(SG必为非0)。
其次,如果有大于1堆的石子的数量大于2,那么如果SG为非0,那么先手只要将SG变为0,后手无论进行什么操作,SG都会变成非0(异或的性质),这样进行下去,终会达到必胜态(只有一堆石子的最大数量大于1)。
[code]#include<iostream> #include <algorithm> #include <cstdio> #include <string.h> using namespace std; long long a[10005]; int main() { #ifndef ONLINE_JUDGE //freopen("1.txt", "r", stdin); #endif int n, i, j, t; long long ans; cin >> t; bool flag; while(t--) { flag = false; cin >> n; ans = 0; for (i = 0; i < n; i++) { cin >> a[i]; ans ^= a[i]; if (a[i] > 1) { flag = true; } } if (flag) { if (ans) { cout << "1\n"; } else { cout << "2\n"; } } else { if (ans) { cout << "2\n"; } else { cout << "1\n"; } } } return 0; }
相关文章推荐
- AngularJs 构建复杂应用(一)
- [IOS]CoreAnimation基础[翻译4部分]
- js流程控制语句(switch语句)
- Java相关书籍推荐
- 几款国产开源的Windows界面库
- Apache Spark 入门简介
- git图解:代码区域总结
- 数据库 mysql 优化器原理
- eclipse导出可执行的jar包
- 友谊赛 排序 水题
- Linux系统负载
- 最大流:Dinic模板
- UVA - 1252 Twenty Questions (状压dp)
- Wolf and Rabbit
- 深入理解JavaScript系列(2):揭秘命名函数表达式
- 共享内存区
- JAVA UDP聊天室
- 七个例子帮你更好地理解 CPU 缓存
- 关于div的居中显示
- HDU5409---CRB and Graph 2015多校 双联通分量缩点