ACM——Quicksum

Quicksum

时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte
总提交:615            测试通过:256

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 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

 

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.

 

输出

 

For each packet, output its Quicksum on a separate line in the output.

 

样例输入

ACM
MID CENTRAL
REGIONAL PROGRAMMING CONTEST
ACN
A C M
ABC
BBC
#

样例输出

46
650
4690
49
75
14
15

提示

 

 

 

题目来源

Mid-Central USA Region 2006

http://acm.njupt.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1027

 

 1 #include<iostream>
 2 #include<string>
 3 #include<fstream>
 4 using namespace std;
 5 int main()
 6 {
 7     string s;
 8     ifstream in("./data.txt");
 9     while(getline(in,s)){
10     cout<<s<<endl;
11     if(s=="#") return 0;
12     int sum=0;
13     for(size_t i=0;i<s.length();i++)
14     {
15         if(s[i]!=' '){
16         sum+=(i+1)*(s[i]-64);
17         }
18     }
19     cout<<sum<<endl;
20     }
21     return 0;
22 }

 

 

 
posted @ 2014-07-21 13:49  123木头人啊  阅读(489)  评论(0编辑  收藏  举报