zoukankan      html  css  js  c++  java
  • 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;
    }
  • 相关阅读:
    1.文件I/O
    sqlite-按日期分组,根据日期查询详细内容
    sqlite-在数据库中创建默认时间
    Git-git 忽略 IntelliJ .idea文件
    重启猫(modem)的方法
    从TP、FP、TN、FN到ROC曲线、miss rate、行人检测评估
    畅所欲言第1期
    使用属性表:VS2013上配置OpenCV
    关于OOM那些事儿
    深度学习之江湖~那些大神们
  • 原文地址:https://www.cnblogs.com/crazyacking/p/3762039.html
Copyright © 2011-2022 走看看