HDU------checksum

Quicksum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3401 Accepted Submission(s): 2095

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

 

Source

Mid-Central USA 2006

 

Recommend

teddy

 

 #include <string.h>
 #include <stdio.h>

 int main()
 {
     char str[];
     int len,i,a[],sum;
     for(i = ;i<;i++)
     {
         a[i] = i+;
     }
     while(gets(str))
     {
         if(strcmp(str,"#") == )
         break;
         len = strlen(str);
         sum = ;
         for(i = ;i<len;i++)
         {
             if(str[i]>='A' && str[i]<='Z')
             sum+=(i+)*a[str[i]-'A'];
         }
         printf("%d\n",sum);
     }

     return ;
 }