codeforces 167C Wizards and Numbers(找规律，博弈)

怼烎@ 2022-06-05 00:50 206阅读 0赞

C. Wizards and Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In some country live wizards. They love playing with numbers.

The blackboard has two numbers written on it — *a* and *b*. The order of the numbers is not important. Let's consider *a* ≤ *b* for the sake of definiteness. The players can cast one of the two spells in turns:

*  Replace *b* with *b* - *a**k*. Number *k* can be chosen by the player, considering the limitations that *k* > 0 and *b* - *a**k* ≥ 0. Number *k* is chosen independently each time an active player casts a spell.
 *  Replace *b* with *b mod a*.

If *a* > *b*, similar moves are possible.

If at least one of the numbers equals zero, a player can't make a move, because taking a remainder modulo zero is considered somewhat uncivilized, and it is far too boring to subtract a zero. The player who cannot make a move, loses.

To perform well in the magic totalizator, you need to learn to quickly determine which player wins, if both wizards play optimally: the one that moves first or the one that moves second.

Input

The first line contains a single integer *t* — the number of input data sets (1 ≤ *t* ≤ 104). Each of the next *t* lines contains two integers *a*, *b*(0 ≤ *a*, *b* ≤ 1018). The numbers are separated by a space.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecificator.

Output

For any of the *t* input sets print "First" (without the quotes) if the player who moves first wins. Print "Second" (without the quotes) if the player who moves second wins. Print the answers to different data sets on different lines in the order in which they are given in the input.

Examples

input

4
    10 21
    31 10
    0 1
    10 30

output

First
    Second
    Second
    First

Note

In the first sample, the first player should go to (11,10). Then, after a single move of the second player to (1,10), he will take 10 modulo 1 and win.

In the second sample the first player has two moves to (1,10) and (21,10). After both moves the second player can win.

In the third sample, the first player has no moves.

In the fourth sample, the first player wins in one move, taking 30 modulo 10.

题意：

给你两个数，两个人轮流进行操作，每轮有两个操作，假设当前这一轮b>a，那么可以把b变成b%a或者是b-a^k。改变后的数不能小于0，当谁不能操作谁就输了。

思路：

每轮操作都可以将(a,b)变成(b%a,a)，如果(b%a,a)是必败态，那么(a,b)是必胜态。如果(b%a,a)是必胜态，那么我们可以打表找规律，会发现周期为a+1。

AC代码：

#include <iostream>  
    #include <cstdio>  
    #include <cstring>  
    #include <string>  
    #include <cstdlib>  
    #include <cmath>  
    #include <vector>  
    #include <queue>  
    #include <map>  
    #include <algorithm>  
    #include <set>  
    #include <functional>  
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    const int INF = 1e9 + 5;
    const int MAXN = 100000007;
    const int MOD = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1.0);
    LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a%b); }
    
    LL getSG(LL x, LL y)
    {
    	if (x == 0 || y == 0)
    		return 0;
    	LL a = min(x, y);
    	LL b = max(x, y);
    	if (getSG(b%a, a))
    	{
    		b /= a;
    		return !((b % (a + 1)) & 1);
    	}
    	return 1;
    }
    
    int main()
    {
    	LL a, b;
    	int T;
    	scanf("%d", &T);
    	while (T--)
    	{
    		scanf("%lld%lld", &a, &b);
    		if (getSG(a, b) == 0)
    			printf("Second\n");
    		else
    			printf("First\n");
    	}
    }