题目描述

Er Ba Game is a popular card game in China’s Province Zhejiang.

In this problem, the game involves two players and 36 cards - 2,3,⋯ ,102,3,\cdots, 102,3,⋯,10, each with four.

In the game each player takes 2 cards (a1,b1)(a_1,b_1)(a1,b1) , (a2,b2)(a_2,b_2)(a2,b2) .Player i{i}i gets (ai,bi)(a_i,b_i)(ai,bi) . Suppose a1≤b1,a2≤b2a_1\leq b_1,a_2\leq b_2a1≤b1,a2≤b2, and we determine the winner as follow:

(2,8){(2,8)}(2,8) is the biggest pair.
If neither is (2,8){(2,8)}(2,8), a pair with a=b{a=b}a=b is greater than a pair without.
If a1=b1a_1=b_1a1=b1 and a2=b2a_2=b_2a2=b2, then compare a1,a2a_1,a_2a1,a2.The bigger one wins.
If a1≠b1,a2≠b2a_1\ne b_1,a_2\ne b_2a1\=b1,a2\=b2, compare (a1+b1) mod 10,(a2+b2) mod 10(a_1+b_1)\ \text{mod}\ 10,(a_2+b_2)\ \text{mod}\ 10(a1+b1) mod 10,(a2+b2) mod 10.The bigger one wins.
If a1≠b1,a2≠b2a_1\ne b_1,a_2\ne b_2a1\=b1,a2\=b2 and (a1+b1) mod 10=(a2+b2) mod 10(a_1+b_1)\ \text{mod}\ 10=(a_2+b_2)\ \text{mod}\ 10(a1+b1) mod 10=(a2+b2) mod 10,compare b1,b2b_1,b_2b1,b2.The bigger one wins.

If a1≤b1,a2≤b2a_1\leq b_1,a_2\leq b_2a1≤b1,a2≤b2 does not hold, we can exchange a1a_1a1 and b1b_1b1 or exchange a2a_2a2 and b2b_2b2 until the formula holds.

You’re told a1,b1,a2,b2a_1,b_1,a_2,b_2a1,b1,a2,b2, and you should tell who’s gonna win the game.

输入描述:

The first line contains an integer T{T}T — number of game cases.
Then T{T}T lines, each contains four integers a1,b1,a2,b2a_1,b_1, a_2,b_2a1,b1,a2,b2, denoting the cards of each player.

输出描述:

Output T{T}T lines.


If player 1 wins, output "first" (without quote).


If player 2 wins, output "second" (without quote).


Otherwise, output "tie" (without quote).

示例1

输入

复制

输出

复制

first
first
second
second
tie

说明

The answers to the first four sets of data are obtained through the first, second, second, and fifth rules respectively.

备注:

It's guaranteed that T≤100,2≤a1,b1,a2,b2≤10T\leq 100,2\leq a_1,b_1,a_2,b_2\leq 10T≤100,2≤a1,b1,a2,b2≤10.

It's not guaranteed that a1≤b1,a2≤b2a_1\leq b_1,a_2\leq b_2a1≤b1,a2≤b2.

思路与代码

简单模拟看懂题目后按照题目意思写就行

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb          push_back
#define ppb         pop_back
#define lbnd        lower_bound
#define ubnd        upper_bound
#define endl        '\n'
#define mll         map<ll,ll>
#define msl         map<string,ll>
#define mls         map<ll, string>
#define rep(i,a,b)    for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)
#define trav(a, x)  for(auto& a : x)
#define pll         pair<ll,ll>
#define vl          vector<ll>
#define vll         vector<pair<ll, ll>>
#define vs          vector<string>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define DEBUG       cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x)  trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ini(a)      memset(a,0,sizeof(a))
#define ini2(a,b)   memset(a,b,sizeof(a))
#define rep(i,a,b)  for(int i=a;i<=b;i++)
#define sc          ll T;cin>>T;for(ll Q=1;Q<=T;Q++)
#define lowbit(x)   x&(-x)
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \
    (void)(cout << "L" << __LINE__ \
    << ": " << #x << " = " << (x) << '\n')
#define TIE \
    cin.tie(0);cout.tie(0);\
    ios::sync_with_stdio(false);

//using namespace __gnu_pbds;

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int maxn = 2000;
const int N = 500009;


void solve(){
	int a1,b1,a2,b2,ans = 0;
	cin>>a1>>b1>>a2>>b2;
	if(a1>b1)a1^=b1^=a1^=b1;
	if(a2>b2)a2^=b2^=a2^=b2;
	if(a1==a2&&b1==b2){//相等 
		ans = 0;
	}else if(a1==2&&b1==8){
		ans = 1;
	}else if(a2==2&&b2==8){
		ans = 2;
	}else if(a1==b1&&a2==b2){
		if(a1>a2){
			ans = 1;
		}else if(a1<a2){
			ans = 2;
		}else{
			ans = 0;
		}
	}else if(a1==b1){
		ans = 1;
	}else if(a2==b2){
		ans = 2;
	}else if(a1!=b1&&a2!=b2){
		int q1 = (a1+b1)%10;
		int q2 = (a2+b2)%10;
		if(q1>q2)ans = 1;
		else if(q1<q2)ans = 2;
		else{
			if(b1>b2)ans = 1;

			else if(b1<b2)ans = 2;
			else ans = 0;
		}
	}else{
		ans = 0;
	}
	if(ans==0){
		cout<<"tie"<<endl;
	}else if(ans==1){
		cout<<"first"<<endl;
	}else{
		cout<<"second"<<endl;
	}
}

int main()
{
//    #ifndef ONLINE_JUDGE
//    freopen ("input.txt","r",stdin);
//    #else
//    #endif
//	solve();
    sc{solve();}
//    int p;
//    while(cin>>p){
//    	while(p--){
//    		solve();
//		}
//	}
//    sc{cout<<"Case "<<Q<<":"<<endl;solve();}
}