hdoj 2734 Quicksum(字符串)

Quicksum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3162    Accepted Submission(s): 2328

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
#
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    代码：
   
   
   
   
   
   
   
   

    
    
    
    
#include<stdio.h>
#include<string.h>
 int main()
 {
 	char str[300];//str[]数组 
 	int sum;
 	while(gets(str))
 	{
 		
 		if(str[0]=='#')
 			break; 
 		else
 		{   sum=0;
 			int l=strlen(str);
 			for(int i=0;i<l;i++)
 			{
 				if(str[i]==' ')//哎 一开始少了一个=，浪费时间 
 				{
 					sum+=0;
				 }
				 else
				 {
				 	sum+=(i+1)*(str[i]-'A'+1);
				 }
			 }
			 printf("%d\n",sum);
		 }
		 
	 }
	 return 0;
 }
    
    
    
    

    
    
    
    

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    46
650
4690
49
75
14
15
   
   
   
   

 

Source

Mid-Central USA 2006