杭电ACM题 1036 Edge解题报告

Edge

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1157 Accepted: 811

Description

For products that are wrapped in small packings it is necessary that the sheet of paper containing the directions for use is folded until its size becomes small enough. We assume that a sheet of paper is rectangular and only folded along lines parallel to its initially shorter edge. The act of folding along such a line, however, can be performed in two directions: either the surface on the top of the sheet is brought together, or the surface on its bottom. In both cases the two parts of the rectangle that are separated by the folding line are laid together neatly and we ignore any differences in thickness of the resulting folded sheet.

After several such folding steps have been performed we may unfold the sheet again and take a look at its longer edge holding the sheet so that it appears as a one-dimensional curve, actually a concatenation of line segments. If we move along this curve in a fixed direction we can classify every place where the sheet was folded as either type A meaning a clockwise turn or type V meaning a counter-clockwise turn. Given such a sequence of classifications, produce a drawing of the longer edge of the sheet assuming 90 degree turns at equidistant places.

Input

The input contains several test cases, each on a separate line. Each line contains a nonempty string of characters A and V describing the longer edge of the sheet. You may assume that the length of the string is less than 200. The input file terminates immediately after the last test case.

Output

For each test case generate a PostScript drawing of the edge with commands placed on separate lines. Start every drawing at the coordinates (300,420) with the command "300 420 moveto". The first turn occurs at (310,420) using the command "310 420 lineto". Continue with clockwise or counter-clockwise turns according to the input string, using a sequence of "x y lineto" commands with the appropriate coordinates. The turning points are separated at a distance of 10 units. Do not forget the end point of the edge and finish each test case by the commands stroke and showpage.

You may display such drawings with the gv PostScript interpreter, optionally after a conversion using the ps2ps utility.

Sample Input

VAVV

Sample Output

300 420 moveto310 420 lineto310 430 linetostrokeshowpage300 420 moveto310 420 lineto310 410 lineto320 410 lineto320 420 linetostrokeshowpage

Source

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

这题不难，就是题目看的比较蛋疼，其实就是当输入A时右转，V的时候左转，每次转后需要走10单位。

我设置了一个方向direc变量，用于记录当前的位置，用x和y记录坐标。我提交两次，一次RE，修改了string数组的大小，本来是20，改成200后就行了。

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

AC代码：

/************************   程序名:Edge.cpp*   功能:ACM************************/#include <iostream>#include <Cstring>using namespace std;#define start_x 300#define start_y 420#define move 10int main(){    char c, string[200];    int length, i, x, y;    char direc;    while(cin>>string) {        direc = 'r';        x = start_x + move;        y = start_y;        length = strlen(string);        cout<<start_x<<" "<<start_y<<" "<<"moveto"<<endl;        cout<<start_x+move<<" "<<start_y<<" "<<"lineto"<<endl;        for(i = 0; i < length; i++) {            if(string[i] == 'V') {                if(direc == 'r') {                    direc = 'u';                    y += move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else if(direc == 'l'){                    direc = 'd';                    y -= move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else if(direc == 'u') {                    direc = 'l';                    x -= move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else {                    direc = 'r';                    x += move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }            }            else if(string[i] == 'A') {                if(direc == 'r') {                    direc = 'd';                    y -= move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else if(direc == 'l') {                    direc = 'u';                    y += move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else if(direc == 'u') {                    direc = 'r';                    x += move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }                else {                    direc = 'l';                    x -= move;                    cout<<x<<" "<<y<<" "<<"lineto"<<endl;                }            }        }        cout<<"stroke"<<endl<<"showpage"<<endl;    }    return 0;}