链接：https://ac.nowcoder.com/acm/contest/11253/D 来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 262144K，其他语言524288K 64bit IO Format: %lld
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:
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(2,8){(2,8)}(2,8) is the biggest pair.
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If neither is (2,8){(2,8)}(2,8), a pair with a=b{a=b}a=b is greater than a pair without.
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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.
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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.
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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
5
2 8 4 6
6 6 6 7
4 5 5 5
6 3 9 10
7 2 2 7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();}
}