URAL 2068 Game of Nuts (博弈)

题意：给定 n 堆石子，每次一个人把它们分成三堆都是奇数的，谁先不能分，谁输。

析：因为每堆都是奇数，那么最后肯定都是要分成1的，那么就把不是1的全加和，然后判断奇偶就OK了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define frer freopen("in.txt", "r", stdin)

#define frew freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 8;

const int mod = 1e9 + 7;

const char *mark = "+-*";

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int main(){

    while(scanf("%d", &n) == 1){

        int ans = 0;

        for(int i = 0; i < n; ++i){

            scanf("%d", &m);

            if(m > 2)

                ans += m/2;

        }

        if(ans % 2 == 0)  puts("Stannis");

        else  puts("Daenerys");

    }

    return 0;

}