题解:
我们可以看到数据量并不是很大,所以我们可以选择一秒钟一秒钟来对这个比赛进行分析
在每一秒中
要判断
1.是否有人到达终点
2.小明与小红之间的距离

/*Keep on going Never give up*/
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
const int maxn = 1e6+10;
const int MaxN = 0x3f3f3f3f;
const int MinN = 0xc0c0c00c;
typedef long long ll;
const int mod = 100000000;
using namespace std;


int main()
{
    int va,vr,t,s,l;
    cin>>va>>vr>>t>>s>>l;
    int ming=0,hong=0;
    int cnt=0;
    while(true){
        if(hong>=l&&ming>=l){
            cout<<"Tie "<<cnt<<endl;
            break;
        }
        else if(hong>=l&&ming<l){
            cout<<"Hong "<<cnt<<endl;
            break;
        }
        else if(hong<l&&ming>=l){
            cout<<"Ming "<<cnt<<endl;
            break;
        }
        if(ming-hong>=t){
            int zz=0;
            while(zz<s){
                zz++;
                cnt++;
                hong+=vr;
                if(hong>=l){
                    cout<<"Hong "<<cnt<<endl;
                    return 0;
                }
            }
        }
        cnt++;
        hong+=vr;
        ming+=va;

    }
    return 0;
}