POJ 2083-Fractal

POJ 2083-Fractal

此题与上一题类似，一次AC;
由于控制台屏幕太小 这次我将结果保存在文件中；

#include < iostream >
#include < cmath >
#include < algorithm >
#include < cstdlib >
#include < cstdio >
#include < fstream >
using   namespace  std;

char  mymap[ 2500 ][ 2500 ];

int  leftdot;
int  rightdot;
int  topdot;
int  bottomdot;


void  figure( int  x, int  y, int  degree)
{
    leftdot=min(leftdot,y);
    rightdot=max(rightdot,y);
    topdot=min(topdot,x);
    bottomdot=max(bottomdot,x);


    if(degree==1)
    {
        mymap[x][y]='X';
    }
    else
    {

        int dis=(int)pow((double)3,degree-2)*2;
        figure(x,y,degree-1);
        figure(x,y+dis,degree-1);
        figure(x+dis,y,degree-1);
        figure(x+dis/2,y+dis/2,degree-1);
        figure(x+dis,y+dis,degree-1);
    }



}


int  main()
{

    int n;
    int i,j;
    ofstream file;
    file.open("test.txt");
    while(scanf("%d",&n))
    {
        rightdot=-100000000;
        leftdot=100000000;
        topdot=1000000000;
        bottomdot=-1000000000;
        memset(mymap,' ',sizeof(mymap));
        if(n==-1)
            break;
        figure(1,1,n);
        for(i=leftdot;i<=rightdot;i++)
        {

            for(j=topdot;j<=bottomdot;j++)
                file<<mymap[i][j];
            file<<endl;
        }
        file<<'-'<<endl;
    }
    file.close();
    return 0;
}

答案请见： /Files/abilitytao/test.txt