zoukankan      html  css  js  c++  java
  • ECNU 1002 IP Address

    ECNU 1002 IP Address

    链接

    https://acm.ecnu.edu.cn/problem/1002

    题目

    单点时限: 2.0 sec

    内存限制: 256 MB

    Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of s and s (bits) to a dotted decimal format. A dotted decimal format for an IP address is form by grouping 8 bits at a time and converting the binary representation to decimal representation. Any 8 bits is a valid part of an IP address.10

    To convert binary numbers to decimal numbers remember that both are positional numerical systems, where the first 8 positions of the binary systems are:

    输入格式
    The input will have a number N(1-9) in its first line representing the number of streams to convert. lines will follow.

    输出格式
    The output must have N lines with a doted decimal IP address. A dotted decimal IP address is formed by grouping 8 bit at the time and converting the binary representation to decimal representation.

    样例
    input
    4
    00000000000000000000000000000000
    00000011100000001111111111111111
    11001011100001001110010110000000
    01010000000100000000000000000001
    output
    0.0.0.0
    3.128.255.255
    203.132.229.128
    80.16.0.1

    思路

    英文题目,不是太难,就是把给的字符串转化为ip地址再输出。
    这里采用了一个tag用于标记权值,12864往后,flag用于标记属于第几块,这道题发散一下也就是一道进制转换的题目。
    当tag到8的时候(还未计算),就代表当前这一段结束了,换下一段并重新计算权值。

    代码

      public static void fun() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
          String str = sc.next();
          int[] a = new int[4];
          StringBuffer sb = new StringBuffer(str);
          int[] weight = new int[]{128, 64, 32, 16, 8, 4, 2, 1};
          int flag = 0;
          int tag = 0;
          for (int j = 0; j < 32; j++) {
            int temp = 1;
            if ((sb.charAt(j) == '0')) {
              temp = 0;
            }
            a[flag] += weight[tag] * temp;
            tag++;
            if (tag == 8) {
              tag = 0;
              flag++;
            }
          }
          System.out.println(a[0] + "." + a[1] + "." + a[2] + "." + a[3]);
        }
      }
    
  • 相关阅读:
    前端-浅谈Flex布局
    css-渐变简约的登录设计
    小程序-小程序后台原生图片识别
    小程序-云数据库实现好看的上传文件动态
    小程序-利用云开发操作云数据库实现点赞评论案例
    小程序-云存储实现对文件的上传下载
    小程序-浅谈云函数获取数据和云数据库api获取数据的区别
    小程序-简易加法教你如何使用云函数
    小程序-云数据库的add,get,remove,update
    小程序-你不得不知的Promise封装请求
  • 原文地址:https://www.cnblogs.com/blogxjc/p/14308281.html
Copyright © 2011-2022 走看看