题目链接：http://codeforces.com/problemset/problem/549/C
C. The Game Of Parity
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
There are n cities in Westeros. The i-th city is inhabited by ai people. Daenerys and Stannis play the following game: in one single move, a player chooses a certain town and burns it to the ground. Thus all its residents, sadly, die. Stannis starts the game. The game ends when Westeros has exactly k cities left.
The prophecy says that if the total number of surviving residents is even, then Daenerys wins: Stannis gets beheaded, and Daenerys rises on the Iron Throne. If the total number of surviving residents is odd, Stannis wins and everything goes in the completely opposite way.
Lord Petyr Baelish wants to know which candidates to the throne he should support, and therefore he wonders, which one of them has a winning strategy. Answer to this question of Lord Baelish and maybe you will become the next Lord of Harrenholl.
Input
The first line contains two positive space-separated integers, n and k (1 ≤ k ≤ n ≤ 2·105) — the initial number of cities in Westeros and the number of cities at which the game ends.
The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 106), which represent the population of each city in Westeros.
Output
Print string "Daenerys" (without the quotes), if Daenerys wins and "Stannis" (without the quotes), if Stannis wins.
Examples
input
3 1
1 2 1
output
Stannis
input
3 1
2 2 1
output
Daenerys
input
6 3
5 20 12 7 14 101
output
Stannis
Note
In the first sample Stannis will use his move to burn a city with two people and Daenerys will be forced to burn a city with one resident. The only survivor city will have one resident left, that is, the total sum is odd, and thus Stannis wins.
In the second sample, if Stannis burns a city with two people, Daenerys burns the city with one resident, or vice versa. In any case, the last remaining city will be inhabited by two people, that is, the total sum is even, and hence Daenerys wins.
题解：
1.如果k==n，直接得出结果。
2.如果后手能把奇数城毁完，那么后手必胜（因为剩下的，不管怎么毁，总和都为偶数）。
3.如果后手不能把奇数城毁完，那么：
3.1. 如果倒数第二步操作的人不能把偶数城毁完，那么最后一步操作的人必胜，因为在最后一步时，既有奇数，也有偶数，可以随意调控。
3.2.如果倒数第二步操作的人能把偶数城毁完，那么剩下的就只有奇数城（或0），所以最后只需看剩下的奇数城有几座，即k。
代码如下：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 2e5+10;
int n, k, a[2];
void init()
{
scanf("%d%d",&n,&k);
a[0] = a[1] = 0;
for(int i = 1; i<=n; i++)
{
int x;
scanf("%d",&x);
a[x%2]++;
}
}
void solve()
{
int winner;
int step = n - k;
if(step==0)
winner = a[1]%2;
else if(step/2>=a[1])
winner = 0;
else
{
if(step/2<a[0]) winner = step%2;
else winner = k%2;
}
if(winner) puts("Stannis");
else puts("Daenerys");
}
int main()
{
init();
solve();
}
