逻辑题目！都来讨论哈

/*

 *环行跑道周长为400米.甲乙两人同时同地顺时针沿环行跑道跑步.甲52米/分钟.

 *乙46米/分钟.假设甲乙两人每跑100米停下休息一分钟.问甲多长时间追上乙? 

 */



struct Accuracy_Control

{

    static const long double time_acc = 1E-2;

    static const long double position_acc = 5.0E-3;

};



struct Racetrack

{

    static const long double circumference = 400.0L;

};



struct Runner

{

    bool is_running;

    long double speed;

    long double position;

    long double status_remain_time;

    long double time;

    

    Runner():

        is_running(false),

        speed(0.0L),

        position(0.0L),

        status_remain_time(0.0L),

        time(0.0L)

    {}

/×

    friend bool operator > ( const Runner& lhs, const Runner& rhs )

    {

        return lhs.position > rhs.position;

    }×/

};



bool need_change_tactic( const Runner& runner, const Accuracy_Control& ac )

{

    return runner.status_remain_time < ac.time_acc;

}



template< typename Run_Tactic, typename Rest_Tactic >

void change_tactic( Runner& runner, const Run_Tactic& run_tactic, 

                           const Rest_Tactic& rest_tactic, const Accuracy_Control& ac )

{

    if ( runner.is_running )

        rest_tactic( runner);

    else

        run_tactic( runner, ac );

}



void time_flying( Runner& runner, const Accuracy_Control& ac )

{

    runner.position += runner.speed;

    runner.status_remain_time -= ac.time_acc;

    runner.time += ac.time_acc;

}



void correct_position( Runner& runner, const Racetrack& racetrack )

{

        if ( runner.position > racetrack.circumference )

        {

            runner.position -= racetrack.circumference;

            //last_relation = !last_relation;

        }

}



bool do_they_meet( const Runner& runner_a, const Runner& runner_b, const Accuracy_Control& ac )

//bool do_they_meet( const Runner& runner_a, const Runner& runner_b, const Accuracy_Control& )

{

    /*

    static bool last_relation = true;

    bool ans = false;

    

    if ( ( runner_a > runner_b ) != last_relation )

        ans = true;



    last_relation = runner_a > runner_b;



    return ans;

    */

    

    bool ans = false;

    

    long double difference = runner_a.position - runner_b.position;

    if ( difference < 0 ) 

        difference = -difference;

    if ( difference < ac.position_acc )

        ans = true;



    return ans;

    

}



struct Run_Tactic_A

{

    void operator()( Runner& runner, const Accuracy_Control& ac ) const

    {

        runner.is_running = true;

        runner.speed = 52.0L / 60.0L * ac.time_acc;

        runner.status_remain_time = 100.0L / 52.0L;

    }

};



struct Run_Tactic_B

{

    void operator()( Runner& runner, const Accuracy_Control& ac ) const

    {

        runner.is_running = true;

        runner.speed = 46.0L / 60.0L * ac.time_acc;

        runner.status_remain_time = 100.0L/ 46.0L;

    }

};



struct Rest_Tactic

{

    void operator()( Runner& runner ) const

    {

        runner.is_running = false;

        runner.speed = 0.0L;

        runner.status_remain_time = 60.0L;

    }

};





#include <iostream>

using namespace std;



int main()

{

    Accuracy_Control ac;

    Racetrack racetrack;

    Runner runner_a, runner_b;



    for ( long long l = 0; l < 99999999; ++l )

    {

        if ( need_change_tactic( runner_a, ac ) )

        {

           change_tactic( runner_a, Run_Tactic_A(), Rest_Tactic(), ac ); 

        }

        if ( need_change_tactic( runner_b, ac ) )

        {

           change_tactic( runner_b, Run_Tactic_B(), Rest_Tactic(), ac ); 

        }

        

        time_flying( runner_a, ac );

        time_flying( runner_b, ac );



        correct_position( runner_a, racetrack );

        correct_position( runner_b, racetrack );



        if ( do_they_meet( runner_a, runner_b, ac ) )

        {

            cout << "\nA. Find a meet at the time " << runner_a.time 

                 << " at the position " << runner_a.position;

            cout << "\nB. Find a meet at the time " << runner_b.time 

                 << " at the position " << runner_b.position;

        }

    }



    return 0;

}

假设甲用t分钟追上乙 此时甲行驶了(x百米+y米)      其中y<100

则甲实际跑了 (t-x) 分钟         总路程为 100*x+y

  乙实际跑了 [t-(x-4)] 分钟     总路程为 100*(x-4)+ y



从而可以列出下面两个等式 :



100*x+y = 52*(t-x)



100*(x-4)+y = 46*[t-(x-4)]



从中可以求得 t-x = 292/3 (甲实际跑的时间)



甲实际跑的路程为 292/3 * 52 = 15184/3  (整数部分为5061)



所以甲中途休息了 50分钟



总的时间为 292/3 + 50 =  147+ 1/3