Codeforces Round #114 (Div. 1) C. Wizards and Numbers 博弈论

C. Wizards and Numbers

题目连接：

http://codeforces.com/problemset/problem/167/C

Description

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 - ak. Number k can be chosen by the player, considering the limitations that k > 0 and b - ak ≥ 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 %I64d specificator.

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.

Sample Input

4

10 21

31 10

0 1

10 30

Sample Output

First

Second

Second

First

Hint

题意

中文题面见：http://acm.uestc.edu.cn/#/problem/show/1169

题解：

代码

#include<bits/stdc++.h>
using namespace std;

int check(long long a,long long b)
{
    if(a==0)return 0;
    if(check(b%a,a))
    {
        b/=a;
        return !((b%(a+1))&1);
    }
    return 1;
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        long long a,b;
        cin>>a>>b;
        if(a>b)swap(a,b);
        if(check(a,b))cout<<"First"<<endl;
        else cout<<"Second"<<endl;
    }
}