URAL2068：Game of Nuts（博弈）

2068. Game of Nuts

Time limit: 1.0 second

Memory limit: 64 MB

The war for Westeros is still in process, manpower and supplies are coming to an end and the winter is as near as never before. The game of thrones is unpredictable so Daenerys and Stannis decided to
determine the true ruler of Seven Kingdoms playing more predictable and shorter game. For example, the game of nuts is the ideal candidate.

Rules of this game are quite simple. Players are given n heaps of nuts. There is an odd number of nuts in each heap. Players alternate turns. In each turn player chooses an arbitrary heap and
divides it into three non-empty heaps so as there is again an odd number of nuts in each heap. The player who cannot make a move loses.

Daenerys has dragons so she moves first. Your task is to determine the winner assuming both Daenerys and Stannis play optimally. Please, do it and stop the war for Westeros!

Input

In the first line there is an odd integer n (1 ≤ n ≤ 777).

In the second line there are n integers separated by spaces. These are the amounts of nuts in each heap at the beginning of the game. It is guaranteed that each heap contains not less than
one and not more than 54321 nuts and this amount is an odd number.

Output

Output "Daenerys" (without quotes) in case of Daenerys’ win. Otherwise output "Stannis".

Samples

input	output
1 3	Daenerys
3 1 1 1	Stannis
5 777 313 465 99 1	Daenerys

Problem Author: Alexey Danilyuk
Problem Source: Ural Regional School Programming Contest 2015

题意：有n堆数量均为奇数的石子，轮流操作，选择任意1堆将其分成3堆奇数的石子，轮到谁不能分的就输。

思路：可以猜下套路，考虑若初始有1堆时发现（A类）1,5,9,13,17...先手输，（B类）3,7,11,15...后手输，拓展到多堆的情况，若初始N堆均为A类，显然先手输，因为后手采取最优策略；若N堆均为B类，取N个3测试发现B类的堆数为奇数就赢，偶数就输；然后AB类混合的情况取A类的堆的数量均为1，B类堆的数量均为3，发现A类对结果无影响，因此问题变成统计N个堆中A类堆的数量。

# include <iostream>
# include <cstdio>
using namespace std;
int main()
{
int n, k;
while(~scanf("%d",&n))
{
int cnt = 0;
while(n--)
{
scanf("%d",&k);
cnt += k>>1;
}
if(cnt&1)
puts("Daenerys");
else
puts("Stannis");
}
return 0;
}