cf435C 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

又一模拟题。。。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define ll long long
#define inf 0x7fffffff
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x*=10;x+=ch-'0';ch=getchar();}
    return x*f;
}
int n,tot,mx=1000,mn=1000,x,y=1000;
int a[1005];
int b[2005][2005];
int main()
{
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read(),tot+=a[i];
    for(int i=1;i<=n;i++)
    {
        x++;
        if(i&1)
        {
            a[i]--;
            b[x][y]=1;
            while(a[i]--)
            {
                x++;y++;
                b[x][y]=1;
            }
            mx=max(mx,y);
        }
        else
        {
            a[i]--;
            b[x][y]=2;
            while(a[i]--)
            {
                x++;y--;
                b[x][y]=2;
            }
            mn=min(mn,y);
        }
    }
    for(int i=mx;i>=mn;i--)
    {
        for(int j=1;j<=tot;j++)
            if(b[j][i]==1)printf("/");
            else if(b[j][i]==2)printf("\\");
            else printf(" ");
        printf("\n");
    }
    return 0;
}