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C. The Game Of Parity
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputThere 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 output
Stannis input
3 1 output
Daenerys input
6 3 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. 题目大意： n个城市，每人轮流屠掉一个城市，最后剩下k个城市，如果剩下的k个城市的人数和是奇数，那么先手赢，否则后手赢。 思路： 如果后手能够有机会屠掉所有奇数城，那么先手必败。 如果有奇数城剩余，继续考虑如果要屠的城为奇数座，那么先手可以最后利用偶数城，调整当前奇数城的数量，必胜。 但是如果后手能够将偶数城全部干掉，那么最后就变成看剩下的k是奇数还是偶数了。 如果要杀偶数个城，那么后手可以做最后调整，如果先手杀不光偶数城，那么后手必胜，如果先手能杀光偶数城， 那么结果就看剩下的k是奇数还是偶数了。 #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<algorithm> #include<cstring> #include<string> #include<vector> #include<map> #include<set> #include<queue> using namespace std; int n,k,a[200001]; int main() { scanf("%d%d",&n,&k); int i,cnt=0; for (i=1;i<=n;i++) { int x; scanf("%d",&x); if (x&1) cnt++; } int num2=(n-k)>>1; int num1=(n-k+1)>>1; bool fg=true; if (cnt<=num2) fg=false; if (k%2==0&&n-cnt<=num2) fg=false; if ((n-k)%2==0) { if ((num1>=n-cnt&&k%2)==false) fg=false; } if (n==k) fg=(cnt&1)?true:false; puts(fg?"Stannis":"Daenerys"); return 0; }
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