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URAL 2068 Game of Nuts (博弈)

2016-08-22 19:46 381 查看
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;
}


  
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