#include<bits/stdc++.h> #define ll long long using namespace std; int main() { int v1,v2,t,s,l; int timeT=0; int L1=0,L2=0; cin>>v1>>v2>>t>>s>>l; while(L1!=l&&L2!=l) { L1+=v1; L2+=v2; timeT++; if(L1-L2>=5) { L2+=s*v2; timeT+=s; } } if(L1l&&L2!=l) cout<<"R"<<endl; else if(L2l&&L1!=l) cout<<"T"<<endl; else if(L1l&&L2l) cout<<"D"<<endl; cout<<timeT; return 0; }

3 comments

  • @ 2023-8-9 17:10:38
    #include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        // 兔子速度:v1米/秒,领先 t 米或以上就休息 s 秒
        // 乌龟速度:v2米/秒
        // 赛道长度:l 米
        int v1, v2, t, s, l;
        cin >> v1 >> v2 >> t >> s >> l;
        // 当前时间(秒),当前兔子跑了的距离,当前乌龟跑了的距离
        int nowTime, nowR, nowT, resTime;
        nowTime = 0;
        nowR = 0;
        nowT = 0;
        while (nowR < l && nowT < l)
        {
            if (nowR - nowT < t)
            {
                nowR += v1;
                nowT += v2;
                nowTime += 1;
            }
            else
            {
                nowR += 0;
                nowT += v2 * s;
                nowTime += s;
            }
        }
        if (nowR >= l && nowT >= l)
        {
            cout << "D\n";
            cout << nowTime << "\n";
        }
        else if (nowR >= l)
        {
            cout << "R\n";
            cout << nowTime << "\n";
        }
        else if (nowT >= l)
        {
            cout << "T\n";
            cout << l / v2 << "\n";
        }
        return 0;
    }
    
    
    • @ 2023-8-9 17:06:31

      龟兔赛跑预测

      • @ 2023-8-8 16:33:47
        #define ll long long
        using namespace std;
        int main()
        {
        	int v1,v2,t,s,l;
        	int timeT=0;
        	int L1=0,L2=0;
        	cin>>v1>>v2>>t>>s>>l;
        	while(L1!=l&&L2!=l)
        	{
        		L1+=v1;
        		L2+=v2;
        		timeT++;
        		if(L1-L2>=5)
        		{
        			L2+=s*v2;
        			timeT+=s;
        		}
        	}
        	if(L1==l&&L2!=l)  cout<<"R"<<endl;
        	else if(L2==l&&L1!=l) cout<<"T"<<endl;
        	else if(L1==l&&L2==l) cout<<"D"<<endl;
        	cout<<timeT;
        	return 0;
        }`````这里重新发下
        
        • 1