学步园

题意：给出了三种矩阵加密方式，实现

思路：这三种方法加上之前的读数据处理我写伪代码用了一小时左右，有点恶心。。

技巧比较多，初学者可以好好研究一下这个题目的细节处理。不懂的可以给我留言，其实我的代码写的还是有点长了。

错点：忘记了00代表100

统计：212k, 0ms, 2Y

#include <iostream><br /> #include <string><br /> #include <ctype.h><br /> #define F(i,a,b) for (int i=a;i<=b;i++)<br /> using namespace std;<br /> int movel[2][4][2] =<br /> {<br /> { {0, 1}, {1, 0}, {0, -1}, {-1, 0} },<br /> { {1, 0}, {0, 1}, {-1, 0}, {0, -1} }<br /> }, size;<br /> char map[101][101];</p> <p>void shake()<br /> {<br /> int move = 1, next;<br /> F(i,1,size)<br /> {<br /> int now = 1, temp = map[1][i];<br /> F(j,1, size-1)<br /> {<br /> next = (now+move+size)%size;<br /> if (!next)<br /> next = size;<br /> map[now][i] = map[next][i];<br /> now = next;<br /> }<br /> map[now][i] = temp;<br /> move*=-1;<br /> }<br /> }</p> <p>void rattle()<br /> {<br /> int move = -1, next;<br /> F(i, 1, size)<br /> {<br /> int now = 1, temp = map[i][1];<br /> F(j,1,size-1)<br /> {<br /> next = (now+move+size)%size;<br /> if (!next)<br /> next = size;<br /> map[i][now] = map[i][next];<br /> now = next;<br /> }<br /> map[i][now] = temp;<br /> move *= -1;<br /> }<br /> }<br /> void loop()<br /> {<br /> int nextx, nexty;<br /> F(i,1, size/2)<br /> {<br /> int dir = i%2, // dir: 0-counterclock, 1-clock<br /> temp = map[i][i],<br /> moving = 0, nowx = i, nowy = i;<br /> do<br /> {<br /> nextx = nowx+movel[dir][moving][0], nexty = nowy+movel[dir][moving][1];<br /> if ( (nextx == i && nexty == size-i+1 ) ||<br /> (nextx == size-i+1 && nexty == size-i+1 ) ||<br /> (nextx == size-i+1 && nexty == i) )<br /> moving++;<br /> map[nowx][nowy] = map[nextx][nexty];<br /> nowx = nextx, nowy = nexty;<br /> }<br /> while (! (nextx ==i && nexty == i ) );<br /> nowx -= movel[dir][moving][0], nowy -= movel[dir][moving][1];<br /> map[nowx][nowy] = temp;<br /> }<br /> }<br /> int main()<br /> {<br /> string line;<br /> while (getline(cin, line) )<br /> {<br /> // read data<br /> string order = line;<br /> getline(cin, line);<br /> size = (order[0]-48)*10 +order[1]-48;<br /> if (size==0)<br /> size=100;<br /> // build map<br /> int i=1, j=1;<br /> F(k,0, line.length()-1 )<br /> {<br /> map[i][j++] = toupper( line[k] );<br /> if (j>size)<br /> j = 1, i++;<br /> }<br /> if (i<=size)<br /> {<br /> int now = 65;<br /> while (i<=size)<br /> {<br /> map[i][j++] = now++;<br /> if (j>size)<br /> j=1, i++;<br /> if (now>90)<br /> now = 65;<br /> }<br /> }<br /> // solve<br /> F(i,2,order.length()-1)<br /> {<br /> switch (order[i])<br /> {<br /> case 'R':<br /> rattle();<br /> break;<br /> case 'S':<br /> shake();<br /> break;<br /> case 'L':<br /> loop();<br /> }<br /> }<br /> // output<br /> F(i,1,size)<br /> F(j,1,size)<br /> printf("%c", map[i][j]);<br /> printf("/n");<br /> }<br /> return 0;<br /> }<br />