冰淇淋(图形递归

传送门

垃圾佬的冰淇淋

TimeLimit:1000MS  MemoryLimit:128MB

64-bit integer IO format:%lld

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Problem Description

垃圾佬在草稿纸上画了一颗爱心，如下

/\/\
\  /
 \/

垃圾佬发现，3颗爱心可以拼接成一个冰淇淋（吃货），如下

/\/\/\/\
\  /\  /
 \/\/\/
  \  /
   \/

现在呢，垃圾佬想画出第N个冰淇淋（上面的分别为第1个，第2个冰淇淋）。可是他还有别的事情要忙。

SO，剩下的冰淇淋就要你来画了。
当然，报酬是有的。。
每画一个，垃圾佬会给你1000000000000000000000000000 mod 10美元的报酬

Input

一个整数N表示第几个冰淇淋
1<=N<=10。

Output

输出是你画好的冰淇淋。每行末尾没有多余的空格

SampleInput

3

SampleOutput

/\/\/\/\/\/\/\/\
\  /\  /\  /\  /
\/\/\/  \/\/\/
\  /    \  /
\/\/\/\/\/
\  /\  /
\/\/\/
\  /
\/

若要绘制一个n阶的冰淇淋，则需要绘制3个n-1阶的冰淇淋，递归下去即可。注意行末没有空格

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

char picture[1<<12][1<<12];
int bin[13];

void draw(int r,int c,int n)
{
if (n==1)
{
picture[r][c]='/';picture[r][c+1]='\\';picture[r][c+2]='/';picture[r][c+3]='\\';
picture[r+1][c]='\\';picture[r+1][c+3]='/';
picture[r+2][c+1]='\\'; picture[r+2][c+2]='/';
}
else
{
draw(r,c,n-1);
draw(r,c+bin
,n-1);
draw(r+bin[n-1],c+bin[n-1],n-1);
}
}
void print(int n)
{
for (int i=1;i<bin[n+1];++i)
{
for (int j=bin[n+1];j>=1;--j)
{
if (picture[i][j]!=' ')
{
picture[i][j+1]=0;
puts(picture[i]+1);
break;
}
}
}
}
int main()
{
bin[0]=1;
for (int i=1;i<=11;++i)
bin[i]=bin[i-1]*2;
int n;
scanf("%d",&n);
for (int i=0;i<=bin[n+1];++i)
for (int j=0;j<=bin[n+1];++j)
picture[i][j]=' ';
draw(1,1,n);
print(n);
return 0;
}