Codeforces Round #249 (Div. 2) C. Cardiogram

C. Cardiogram

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

In this problem, your task is to use ASCII graphics to paint a cardiogram.

A cardiogram is a polyline with the following corners:



That is, a cardiogram is fully defined by a sequence of positive integers a1, a2, ..., an.

Your task is to paint a cardiogram by given sequence ai.

Input
The first line contains integer n (2 ≤ n ≤ 1000). The next line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 1000). It is guaranteed that the sum of all ai doesn't exceed 1000.

Output
Print max |yi - yj| lines (where yk is the y coordinate of the k-th point of the polyline), in each line print

characters. Each character must equal either « / » (slash), « \ » (backslash), «» (space). The printed image must be the image of the given polyline. Please study the test samples for better understanding of how to print a cardiogram.

Note that in this problem the checker checks your answer taking spaces into consideration. Do not print any extra characters. Remember that the wrong answer to the first pretest doesn't give you a penalty.

Sample test(s)

Input

5
3 1 2 5 1

Output

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

Input

3
1 5 1

Output

 / \
\
\
\
\ / 

Note
Due to the technical reasons the answers for the samples cannot be copied from the statement. We've attached two text documents with the answers below.
http://assets.codeforces.com/rounds/435/1.txt http://assets.codeforces.com/rounds/435/2.txt
//includes
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <utility>
#include <algorithm>
#include <cassert>

using namespace std;

//defines-general
typedef long long ll;
typedef long double ld;
#define to(a) __typeof(a)
#define fill(a,val) memset(a,val,sizeof(a))
#define repi(i,a,b) for(__typeof(b) i = a;i<b;i++)

//defines-pair
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
#define ff first
#define ss second
#define mp make_pair

//defines-vector
typedef vector<int> vi;
typedef vector<long long> vll;
#define all(vec) vec.begin(),vec.end()
#define tr(vec,it) for(__typeof(vec.begin()) it = vec.begin();it!=vec.end();++it)
#define pb push_back
#define sz size()
#define contains(vec,x) (find(vec.begin(),vec.end(),x)!=vec.end())

int main()
{
vector<pii > vals;
char ans[2000][2000];
repi(i, 0, 2000)
{
repi(j, 0, 2000) ans[i][j] = ' ';
}
int n;
cin >> n;
int x=0,y=0;
vals.pb(mp(x,y));
int maxy=0;
int maxx=1;
int add = 1;
repi(i, 0, n)
{
int a;
cin >> a;
x+=a;
y+=((i%2==0)?1:-1)*a;
maxy = max(y,maxy);
maxx = max(x,maxx);
vals.pb(mp(x,y));
}
repi(i, 0, n+1)
{
vals[i].ss = -1*(vals[i].ss-maxy);
}
repi(i, 1, n+1)
{
x = vals[i-1].ff;
y = vals[i-1].ss;
int x1 = vals[i].ff;
int y1 = vals[i].ss;
if(i%2)
{
for(int j = y1+1; j<=y; j++)
{
ans[j-1][x1-(j-y1)] = '/';
}
}
else
{
for(int j = y+1; j<=y1; j++)
{
ans[j-1][x+(j-y-1)] = '\\';
}
}
}
int maxi = 0;
int maxj = 0;
repi(i, 0, 1010)
{
repi(j, 0, 1010)
{
if(ans[i][j]!=' ')
{
maxi = max(i,maxi);
maxj = max(j,maxj);
}
}
}
repi(i, 0, maxi+1)
{
repi(j, 0, maxj+1) printf("%c",ans[i][j]);
cout << endl;
}
// for(pii temp:vals) cout << temp.ff <<" " << temp.ss << endl; cout << endl;
return 0;
}