Quicksum

/*
*程序的版权和版本声明部分：
*Copyright(c)2014，烟台大学计算机学院学生
*All rights reserved.
*文件名称：
*作者：田成琳
*完成日期：2014 年 6 月 12 日
*版本号：v1.0
*对任务及求解方法的描述部分：
*问题描述：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 exampleQuicksum 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

*程序输入: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.

*程序输出：For each packet, output its Quicksum on a separate line in the output.
*问题分析：
*算法设计：
*/
           

题目分析：题中意思说输入案例有多个，每个输入案例中只包含大写字母和空格，然后根据每个大写字母的位置做相应计算。A=1，B=2，C=3...，Z=26。比如ACM：A在第一位，C第二位，M第三位，那么结果就为1*1+2*3+3*13=46。但应引起注意的是案例中有的有空格字符，空格字符看成0对待。那么实现代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    char word[100000];
    int sum,length,i;
    while(cin.getline(word,sizeof(word))&&(string)word!="#")//运用类型转换
    {
        sum=0;
        length=strlen(word);
        for(i=0; i<length; i++)
        {
            while(word[i]==' ')//忽略空格
                i++;
            sum+=(i+1)*(int)(word[i]-64);//类型转换
        }
        cout<<sum<<endl;
    }
    return 0;
}
           

运行结果：