• Codeforces Round #459 (Div. 2) AB


    A. Eleven
     

    Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly n characters.

    Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the i-th letter of her name should be 'O' (uppercase) if i is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to n. Fibonacci sequence is the sequence f where

    • f1 = 1,
    • f2 = 1,
    • fn = fn - 2 + fn - 1 (n > 2).

    As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.

    Input

    The first and only line of input contains an integer n (1 ≤ n ≤ 1000).

    Output

    Print Eleven's new name on the first and only line of output.

    Examples
    input
    8
    output
    OOOoOooO
    input
    15
    output
    OOOoOooOooooOoo

     是斐波拉契数就是‘O’否则就是‘o’

     1 n = int(input())
     2 
     3 vis = []
     4 a ,b = 1, 1
     5 for i in range(55):
     6     vis.append(b)
     7     a,b = b,a+b
     8 s = ""
     9 for i in range(n):
    10     if vis.count(i+1):
    11         s += 'O'
    12     else :
    13         s += 'o'
    14 print(s)
    B. Radio Station
     

    As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx configuration for school's website. The school has n servers. Each server has a name and an ip (names aren't necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we'll assume that an nginx command is of form "command ip;" where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers.

    Each ip is of form "a.b.c.d" where abc and d are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has m commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is "command ip;" Dustin has to replace it with "command ip; #name" where name is the name of the server with ip equal to ip.

    Dustin doesn't know anything about nginx, so he panicked again and his friends asked you to do his task for him.

    Input

    The first line of input contains two integers n and m (1 ≤ n, m ≤ 1000).

    The next n lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1 ≤ |name| ≤ 10, name only consists of English lowercase letters). It is guaranteed that all ip are distinct.

    The next m lines contain the commands in the configuration file. Each line is of form "command ip;" (1 ≤ |command| ≤ 10, commandonly consists of English lowercase letters). It is guaranteed that ip belongs to one of the n school servers.

    Output

    Print m lines, the commands in the configuration file after Dustin did his task.

    Examples
    input
    2 2
    main 192.168.0.2
    replica 192.168.0.1
    block 192.168.0.1;
    proxy 192.168.0.2;
    output
    block 192.168.0.1; #replica
    proxy 192.168.0.2; #main
    input
    3 5
    google 8.8.8.8
    codeforces 212.193.33.27
    server 138.197.64.57
    redirect 138.197.64.57;
    block 8.8.8.8;
    cf 212.193.33.27;
    unblock 8.8.8.8;
    check 138.197.64.57;
    output
    redirect 138.197.64.57; #server
    block 8.8.8.8; #google
    cf 212.193.33.27; #codeforces
    unblock 8.8.8.8; #google
    check 138.197.64.57; #server

     有n台服务器,m条命令,求每个命令的名字指向的服务器ip在n台服务器中原来的名字是什么。

    n,m = map(int,input().split())
    di = {}
    for i in range(n+m):
        s,s1 = input().split()
        if s1[-1] == ';':
            s1 = s1[:-1]
        if s1 in di:
            print(s+' '+s1+'; #'+di[s1])
        else:
            di[s1] = s;
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  • 原文地址:https://www.cnblogs.com/xingkongyihao/p/8387687.html
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