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 #include <stdio.h>

#include <string.h>

#include <iostream>

#include <cmath>

using namespace std;

#define maxn 2010

char g[maxn][maxn];

int main()

{

    memset(g,' ',sizeof(g));

    int num = 0;

    int miny = 1000, maxy = 1000;//最大值最小值都要初始成1000

    int  x = 0, y = 1000,n = 0;

    int d = 1;

    scanf("%d",&n);

    for( int i = 0; i<n; i++)

    {

        scanf("%d",&num);

        while( num--)

        {

            if( d == 1)

              g[y][x++] = '/';

            else

              g[y][x++] = '\\';

            y += d;

        }

        y -= d;//注意每當方向改變時，與上一次最後一個一行；

        d = -d;

        maxy = max(maxy,y);

        miny = min(miny,y);

    }

    for (int i = maxy; i >= miny; --i)

    {

        g[i][x]=0;//加入字元串結束标準，要不然無法用puts

        puts(g[i]);

    }

  //  scanf("%d",&n);

    return 0;

}