HDU-1033-Edge(C++ && 简单模拟 && 题意恶心)

Edge

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2314 Accepted Submission(s): 1477



Problem 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

V
AVV

Sample Output

300 420 moveto
310 420 lineto
310 430 lineto
stroke
showpage
300 420 moveto
310 420 lineto
310 410 lineto
320 410 lineto
320 420 lineto
stroke
showpage

Source

University of Ulm Local Contest 2003

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三天没写，脑袋果然有点小生锈，祝贺我的朋友老赵成为东北赛区ACM正式队员！博主当然是替他高兴啊！无奈水平太差，不能与其共同战斗，所以我呢有时间做做题就好，有句话不是常说，自己选择的路！跪着也要走完，即使我是一次正式比赛也参加过渣渣，但是这份AC之心，永不放弃！

接下来，今天的题目，出题人简直恶心，（明明是个水题非得弄得这么高大上？）我想即使英语很好的人，看这道题都会很费劲吧，更不要说，英语不好的人了（比如说我），说了一圈，题意解释的模模糊糊的，什么叫言简意赅，我也是醉了！里面什么A是顺时针，V是逆时针的，我是看了30分钟都没看懂！

所以这题的题意是：

它给了你一个起点和第一次画线的方向，

第一个起点是(300,420)，第一次画线方向往右画的，

所以每次都得输出：

300 420 moveto

310 420 lineto

然后后面便根据你输入的A,V来重新进行方向画线，

记住A是往右V是往左

然后唯一的难点就是用dir控制东西南北方向的判定了,dir的周期为4

//A往右边，V往左边

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define LEN 300
int main()
{
char cmd[LEN];
while(scanf("%s",cmd)!=EOF)
{
int i, j, len;
int x, y, dir;
dir = 0;
x = 310;
y = 420;
len = strlen(cmd);
printf("300 420 moveto\n");
printf("310 420 lineto\n");
for(i=0;i<len;i++)
{
if('V'==cmd[i])
{
switch(dir)
{
case 0:
y+=10;
break;
case 1:
x+=10;
break;
case 2:
y-=10;
break;
case 3:
x-=10;
break;
}
dir=(dir+3)%4;
printf("%d %d lineto\n",x,y);
}
else if('A'==cmd[i])
{
switch(dir)
{
case 0:
y-=10;
break;
case 1:
x-=10;
break;
case 2:
y+=10;
break;
case 3:
x+=10;
break;
}
dir = (dir+1)%4;
printf("%d %d lineto\n",x,y);
}
}

printf("stroke\n");
printf("showpage\n");
}
}