Codeforces Bayan Bus
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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.
The only line of input contains integer k, (0 ≤ k ≤ 34), denoting the number of participants.
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.
9
+------------------------+|O.O.O.#.#.#.#.#.#.#.#.|D|)|O.O.O.#.#.#.#.#.#.#.#.|.||O.......................||O.O.#.#.#.#.#.#.#.#.#.|.|)+------------------------+
20
+------------------------+|O.O.O.O.O.O.O.#.#.#.#.|D|)|O.O.O.O.O.O.#.#.#.#.#.|.||O.......................||O.O.O.O.O.O.#.#.#.#.#.|.|)+------------------------+#include <iostream>#include <cstdio>using namespace std;char c[6][27];bool vis[6][27]={false};int main(){ int n; cin>>n; c[0][0]='+',c[0][25]='+',c[5][0]='+';c[5][25]='+'; for(int i=1;i<5;i++) { c[i][0]='|'; c[i][25]='|'; } for(int j=1;j<25;j++) { c[0][j]='-'; c[5][j]='-'; } for(int i=1;i<5;i++) for(int j=1;j<24;j++) { if(j%2==0) c[i][j]='.'; else c[i][j]='#'; } for(int i=1;i<5;i++) { c[i][23]='|'; c[i][24]='.'; } c[1][24]='D'; int x,y; if(n<=4) for(int i=1;i<=n;i++) c[i][1]='O'; else { for(int i=1;i<=4;i++) { c[i][1]='O'; vis[i][1]=true; } x=(n-4)/3; y=(n-4)%3; } int p,q=3; for(int i=1;i<=x;i++) { for( p=1;p<5;p++) c[p][q]='O'; q+=2; } for(int i=1;i<=y;i++) { c[i][q]='O'; } for(int j=1;j<25;j++) { if(!vis[3][j]) c[3][j]='.'; } for(int i=0;i<6;i++) for(int j=0;j<26;j++) { cout<<c[i][j]; if((i==1||i==4)&&j==25) cout<<")"; if(j==25) cout<<endl; } return 0;}
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