#651 (Div. 2)C. Number Game（博弈论，思维）

题目描述

Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
Divide n by any of its odd divisors greater than 1.
Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.

Input

The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1≤n≤109).

Output

For each test case, print “Ashishgup” if he wins, and “FastestFinger” otherwise (without quotes).

Example

input
7
1
2
3
4
5
6
12
output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup

Note

In the first test case, n=1, Ashishgup cannot make a move. He loses.
In the second test case, n=2, Ashishgup subtracts 1 on the first move. Now n=1, FastestFinger cannot make a move, so he loses.
In the third test case, n=3, Ashishgup divides by 3 on the first move. Now n=1, FastestFinger cannot make a move, so he loses.
In the last test case, n=12, Ashishgup divides it by 3. Now n=4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.

题目大意

给定一个正整数 n，有两种操作：
（1）用 n 的一个大于 1 的奇因子除 n；（2）如果 n>1 将 n 减 1。
两个人玩游戏，每次每个人可以选择一个操作，最后谁不能操作谁就输。问谁会赢。

题目分析

这是一个博弈论的题目，有两条分支。
1是一个必败态，由1可以推出两个必胜态：2和奇数（这就是两条分支）。

2这条分支：
由2这条分支可以推出，2*一个素数 (素数>2)是一个必败态。（先手除以该素数，会给对手留下2这个必胜态；而-1操作则会给对手留下奇数这个必胜态）
进而可以推出，2 * 一个素数^n (素数>2)是一个必败态。（先手可以直接除以这个素数的n-1次方，留给对手2*一个素数的必败态）
分支1：1(0) -> 2(1) -> 2*素数(0) -> 2*素数^n(1)
奇数这条分支：
由奇数这条分支可以推出，2^n（n>1）是必败态。（因为2ⁿ没法除以奇数了，只能选择-1操作，然后就变成了奇数）
进而可以推出，2^n * 某个奇数是必胜态。（先手可以直接除以这个奇数，然后就给对方留下了2ⁿ的必败态）
分支2：1(0)-> 奇数(1) -> 2^n(0) -> 2^n*奇数(1)

代码如下

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set> 
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=2e3+5;
bool is_prime(int x)
{
	for(int i=2;i<=x/i;i++)
		if(x%i==0) return false;
	return true;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		if(n==1) {puts("FastestFinger"); continue;}
		if(n&1||n==2) {puts("Ashishgup"); continue;}
		bool flag;
		if(n%4==0)	//分支2
		{
			while((n&1)==0) n>>=1;
			if(n==1) flag=false;
			else flag=true;
		}
		else	//分支1
		{
			n>>=1;
			if(is_prime(n)) flag=false;
			else flag=true;
		}
		if(flag) puts("Ashishgup");
		else puts("FastestFinger");
	}
	return 0;
}

来源：oschina

链接：https://my.oschina.net/u/4286638/blog/4320846

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