zoukankan      html  css  js  c++  java
  • hdu-2734

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=2734

    Quicksum

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3531    Accepted Submission(s): 2592


    Problem Description
    A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for detecting transmission errors, validating document contents, and in many other situations where it is necessary to detect undesirable changes in data.

    For this problem, you will implement a checksum algorithm called Quicksum. A Quicksum packet allows only uppercase letters and spaces. It always begins and ends with an uppercase letter. Otherwise, spaces and letters can occur in any combination, including consecutive spaces.

    A Quicksum is the sum of the products of each character's position in the packet times the character's value. A space has a value of zero, while letters have a value equal to their position in the alphabet. So, A=1, B=2, etc., through Z=26. Here are example Quicksum calculations for the packets "ACM" and "MID CENTRAL":

    ACM: 1*1 + 2*3 + 3*13 = 46MID CENTRAL: 1*13 + 2*9 + 3*4 + 4*0 + 5*3 + 6*5 + 7*14 + 8*20 + 9*18 + 10*1 + 11*12 = 650
     
    Input
    The input consists of one or more packets followed by a line containing only # that signals the end of the input. Each packet is on a line by itself, does not begin or end with a space, and contains from 1 to 255 characters.
     
    Output
    For each packet, output its Quicksum on a separate line in the output.

     
    Sample Input
    ACM MID CENTRAL REGIONAL PROGRAMMING CONTEST ACN A C M ABC BBC #
     
    Sample Output
    46 650 4690 49 75 14 15
     
    思路:用字符串输入,然后每个数都减去64再乘以下表加1
     
    难度:简单,就是英文不太看得懂==
     
    代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 int main()
     4 {
     5     int n,i,sum;
     6     char s[255];
     7     while(gets(s))
     8     {
     9         if(s[0]=='#')
    10             break;
    11         sum=0;
    12         n=strlen(s);
    13         for(i=0;i<n;i++)
    14         {
    15             if(s[i]>='A'&&s[i]<='Z')
    16                 sum=sum+(s[i]-64)*(i+1);
    17         }
    18         printf("%d
    ",sum);
    19     }
    20     return 0;
    21 }
  • 相关阅读:
    Python 工程管理及 virtualenv 的迁移
    Python基础系列讲解——random模块随机数的生成
    Python进阶量化交易场外篇5——标记A股市场涨跌周期
    Python学习案例之视频人脸检测识别
    基于python的Splash基本使用和负载均衡配置
    你所听到的技术原理、技术本质到底是什么?
    BAT大厂面试流程剖析
    基于Python的ModbusTCP客户端实现
    互联网寒冬,Python 程序员如何准备面试
    ES-查询后10000条数据的设置
  • 原文地址:https://www.cnblogs.com/ruo786828164/p/5971263.html
Copyright © 2011-2022 走看看