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;
}

 

posted @   北岛知寒  阅读(194)  评论(0编辑  收藏  举报
编辑推荐:
· 如何编写易于单元测试的代码
· 10年+ .NET Coder 心语,封装的思维:从隐藏、稳定开始理解其本质意义
· .NET Core 中如何实现缓存的预热?
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
· AI与.NET技术实操系列:向量存储与相似性搜索在 .NET 中的实现
阅读排行:
· 周边上新:园子的第一款马克杯温暖上架
· Open-Sora 2.0 重磅开源!
· 分享 3 个 .NET 开源的文件压缩处理库,助力快速实现文件压缩解压功能!
· Ollama——大语言模型本地部署的极速利器
· DeepSeek如何颠覆传统软件测试?测试工程师会被淘汰吗?
点击右上角即可分享
微信分享提示
主题色彩