题意：
给定一个正n边形的点。双方轮流连点成线，要求所画的线不能与之前的线相交。当某个人连成一个回路，这个人就输了。问先手必胜还是后手必胜。
思路：
SG函数，因为一条线相当于把图劈成了两半，所以每次用异或运算推过来。
/*
* @Author: chenkexing
* @Date: 2019-01-13 16:17:46
* @Last Modified by: chenkexing
* @Last Modified time: 2019-01-15 18:33:24
*/ #include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert> using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行
#define REP(i , j , k) for(int i = j ; i < k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399; //黄金分割点
const double tPHI=0.38196601; template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
const int maxn = ;
int sg[maxn],s[maxn];
void getsg(int n){ for(int i=; i<=n; i++){
memset(s, , sizeof(s));
for(int j=; j<=i-; j++){
s[(sg[j] ^ sg[i-j-])] = ;
}
for(int j=; ; j++){
if(!s[j]) {
sg[i] = j;
break;
}
}
}
}
int main(){
int T;
getsg();
scanf("%d", &T);
while(T--){
int n; scanf("%d", &n);
if(sg[n])puts("First");
else puts("Second");
} return ;
}