zoukankan      html  css  js  c++  java
  • [字符哈希] POJ 3094 Quicksum

    Quicksum
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 16488   Accepted: 11453

    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 = 46

    MID 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

    Source

     
    原题大意: 一种类似于字符哈希的东西,输入然后输出就好了。
    解题思路:简单模拟。
      
    #include<stdio.h>
    #include<string.h>
    int chash(char *p,int len)
      {
      	 int i,sum=0;
      	 for(i=1;i<=len;++i) if(*p!=' ') sum+=i*(int)((*p++)-'A'+1); else ++p;
      	 return sum;
      }
    int main()
      {
      	 char s[300];
      	 while(gets(s)&&s[0]!='#') printf("%d
    ",chash(s,strlen(s)));
      	 return 0;
      }
    

      

  • 相关阅读:
    语音合成
    JAVA的18条BASE
    Java关键字final、static使用总结
    JAVA学习之路:不走弯路,就是捷径
    每个java初学者都应该搞懂的问题
    Tomcat5.5.9+JSP经典配置实例
    FineUI控件集合
    AngularJS基础
    数据库优化方案之SQL脚本优化
    数据库分库分表策略之MS-SQL读写分离方案
  • 原文地址:https://www.cnblogs.com/fuermowei-sw/p/6250273.html
Copyright © 2011-2022 走看看