CF1370C Number Game (Game Theory + Number Theory)

topic

At the beginning, we can put $n\; is$ divided into three cases

• When $n=1$ , the second hand wins
• When $n$ is odd, win first,nnJust divide$n$ by itself

The first two cases are relatively simple, let's discuss when $When\; n$ is an even number,
when$n$ is an even number (2 2Multiples of$2$ ), if you want to win, you definitely can’t use minus$1$ operation, because the second player will win. Then consider dividing by the divisor
because$n$ is an even number, we can definitely putnndecompose$n$ into some prime numbers and2 2The product of positive integer powers of$2$$n=p_1\times p_2\times \cdots \times p_m\times 2^k$ $(k>0,m\ge 0)$ , and these prime numbers are not2 2A multiple of$2.$
Because the value removed each time can be the product of multiple prime factors (as long as it is a divisor), as long as$m>1$ ,Ashishgupyou can choose the approximate number flexibly.
Finally, when$n$ is removed, only$2^{When\; k}$ , there will be two situations

1. $n=2.$ The person who takes first wins: judge whether the$m$ prime numbers are converted into even numbers that are not$A$ product of the number of, if possible, this descriptionAshishgupcan take the first (Ashishgupwin)
2. $n>2.\; The$ winner wins: judge whether the$m$ prime numbers are converted to odd numbers, not$A$ product of the number of, if possible, this descriptionAshishgupmay be taken after (Ashishgupwinning)

Otherwise, the winner is FastestFingerthe matter

Code

#include<cstdio>
#include<iostream>
using namespace std;
int n;
int main()
{
    
    
//	freopen("in.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--)
	{
    
    
		scanf("%d",&n);
		if(n==1){
    
    puts("FastestFinger");continue;}
		if((n & 1) || n==2){
    
    puts("Ashishgup");continue;}
		int tot=0;
		for(int i=2;i*i<=n;++i)
		if((i & 1) && n%i==0) // 注意这里只能把为奇数的质因数
		while(n%i==0)++tot,n/=i;
		for(int i=2;i*i<=n;++i)
		{
    
    
			int tmp=n/i;
			if((tmp & 1) && n%i==0)
			while(n%tmp==0)++tot,n/=tmp;
		}
		if(n==2)
		{
    
    
			if(tot!=1)puts("Ashishgup");
			else puts("FastestFinger"); // 如果只有一个质数的话就不能化为偶数个数之积
		}
		else
		{
    
    
			if(tot)puts("Ashishgup");
			else puts("FastestFinger"); // 这里特判的是0，跟上面原理相同
		}
	}
	return 0;
}