杭电2056Rectangles(未解决)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2056

该题Output Limit Exceeded，可能是考虑情况太少了！

下面的代码Output Limit Exceeded了

 

View Code

#include <cstdlib>

#include <iostream>

#include <stdio.h>

using namespace std;



int main(int argc, char *argv[])

{

    double a[8]={0};

    //int n=2;

    while(1)

    {

      double x=0,y=0;

      double area=0;

      for(int i=0;i<8;i++)

         cin>>a[i];

      if(a[0]>a[4])

      {

        x=a[6]-a[0];

        y=a[7]-a[1];

        area=x*y;

        printf("%.2lf\n",area);

      }

      if(a[0]<a[4])

      {

        x=a[2]-a[4];

        y=a[3]-a[5];

        area=x*y;

        printf("%.2lf\n",area);

      }

        

    }

    system("PAUSE");

    return EXIT_SUCCESS;

}

 解决它~~~~~~~~~

下面的代码Time Limit Exceeded了

 

View Code

#include <cstdlib>

#include <iostream>

#include <stdio.h>

using namespace std;



int main(int argc, char *argv[])

{

    double a[8]={0};

    //int n=2;

    while(1)

    {

      double x1=0,y1=0,x2=0,y2=0;

      double area=0;

      for(int i=0;i<8;i++)

         cin>>a[i];

      if(a[0]>a[4])

        x1=a[0];

      else

       x1=a[4];

      if(a[1]>a[5])

        y1=a[1];

      else

        y1=a[5];

      if(a[2]>a[6])

        x2=a[6];

      else

        x2=a[2];

      if(a[3]>a[7])

        y2=a[7];

      else

        y2=a[3];

      if(x2>x1&&y2>y1)

      {

        area=(x2-x1)*(y2-y1);

        printf("%.2lf\n",area);

      }

       

        

    }

    system("PAUSE");

    return EXIT_SUCCESS;

}

 

 

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