cf475A Bayan Bus

最新推荐文章于 2014-11-04 09:38:09 发布

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最新推荐文章于 2014-11-04 09:38:09 发布

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原文链接：http://www.cnblogs.com/zhber/p/4035888.html

The final round of Bayan Programming Contest will be held in Tehran, and the participants will be carried around with a yellow bus. The bus has 34 passenger seats: 4 seats in the last row and 3 seats in remaining rows.

The event coordinator has a list of k participants who should be picked up at the airport. When a participant gets on the bus, he will sit in the last row with an empty seat. If there is more than one empty seat in that row, he will take the leftmost one.

In order to keep track of the people who are on the bus, the event coordinator needs a figure showing which seats are going to be taken by k participants. Your task is to draw the figure representing occupied seats.

Input

The only line of input contains integer k, (0 ≤ k ≤ 34), denoting the number of participants.

Output

Print the figure of a bus with k passengers as described in sample tests. Character '#' denotes an empty seat, while 'O' denotes a taken seat. 'D' is the bus driver and other characters in the output are for the purpose of beautifying the figure. Strictly follow the sample test cases output format. Print exactly six lines. Do not output extra space or other characters.

Sample test(s)
input
9
output
+------------------------+
|O.O.O.#.#.#.#.#.#.#.#.|D|)
|O.O.O.#.#.#.#.#.#.#.#.|.|
|O.......................|
|O.O.#.#.#.#.#.#.#.#.#.|.|)
+------------------------+
input
20
output
+------------------------+
|O.O.O.O.O.O.O.#.#.#.#.|D|)
|O.O.O.O.O.O.#.#.#.#.#.|.|
|O.......................|
|O.O.O.O.O.O.#.#.#.#.#.|.|)
+------------------------+

十分的蛋疼……看到样例就吓尿了

唉打了这么久cf手速还是不够快9分钟才A

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
int main()
{
	n=read();
	printf("+------------------------+\n");
	printf("|");
	if (n)printf("O.");else printf("#.");
	for (int i=1;i<=10;i++)
	  if ((n-4)>3*(i-1))printf("O.");else printf("#.");
	printf("|D|)\n");
	if (n>1)printf("|O.");else printf("|#.");
	for (int i=1;i<=10;i++)
	  if ((n-5)>3*(i-1))printf("O.");else printf("#.");
	printf("|.|\n");
	if (n>2)printf("|O");else printf("|#");
	printf(".......................|\n");
	if (n>3)printf("|O.");else printf("|#.");
	for (int i=1;i<=10;i++)
	  if ((n-6)>3*(i-1))printf("O.");else printf("#.");
	printf("|.|)\n");
	printf("+------------------------+\n");
}

转载于:https://www.cnblogs.com/zhber/p/4035888.html