1 solutions

  • 0
    @ 2025-3-3 16:33:28

    C++ :

    #include<iostream>  
      
    using namespace std;  
      
    int main(){  
        int v1,v2,t,s,l;  
        cin>>v1>>v2>>t>>s>>l;  
        int xr=0,xt=0,tr=0,tt=0;  
        while(xr<l && xt<l){  
            if(xr-xt>=t){  
                tt+=s;  
                xt=tt*v2;  
                xr=tr*v1;  
            }  
            else{  
                tr++;  
                tt++;  
                xr=v1*tr;  
                xt=v2*tt;  
            }  
        }  
      
        if(xr==xt){  
            cout<<"D"<<endl;  
            cout<<tt<<endl;  
        }  
        else if(xr>xt){  
            cout<<"R"<<endl;  
            cout<<tt<<endl;  
        }  
        else{  
            cout<<"T"<<endl;  
            cout<<l/v2<<endl;  
        }  
        return 0;  
    }  
    

    Java :

    import java.util.Scanner;
    
    public class Main {
        public static void main(String[] args) {
            int v1, v2, t, s, l, time;
            Scanner sc = new Scanner(System.in);
            v1 = sc.nextInt();
            v2 = sc.nextInt();
            t = sc.nextInt();
            s = sc.nextInt();
            l = sc.nextInt();
            int sum1 = 0, sum2 = 0;
            int flag = 0;
            for (time = 1; time <= 10000; time++) {
                if (sum1 - sum2 >= t && flag == 0) {
                    flag = s;
                }
                if (flag == 0) {
                    // 兔子的路程
                    sum1 += v1;
                } else {
                    flag--;
                }
                // 乌龟的路程
                sum2 += v2;
    
                // 平
                if (sum1 == l && sum2 == l) {
                    System.out.println("D");
                    System.out.println(time);
                    return;
                } else if (sum1 >= l) { // 兔子
                    System.out.println("R");
                    System.out.println(time);
                    return;
                } else if (sum2 >= l) { // 乌龟
                    System.out.println("T");
                    System.out.println(time);
                    return;
                }
            }
        }
    }
    
    
    • 1

    Information

    ID
    10598
    Time
    1000ms
    Memory
    512MiB
    Difficulty
    (None)
    Tags
    # Submissions
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