hdu-2734

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2734

 

Quicksum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3531    Accepted Submission(s): 2592

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

 

思路：用字符串输入，然后每个数都减去64再乘以下表加1

 

难度：简单，就是英文不太看得懂==

 

代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,sum;
    char s[255];
    while(gets(s))
    {
        if(s[0]=='#')
            break;
        sum=0;
        n=strlen(s);
        for(i=0;i<n;i++)
        {
            if(s[i]>='A'&&s[i]<='Z')
                sum=sum+(s[i]-64)*(i+1);
        }
        printf("%d\n",sum);
    }
    return 0;
}