Quicksum

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3516 Accepted Submission(s): 2579

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 = 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

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

题意分析：

水题。每个英文字母在字符串的的位置乘英文字母在字母表中的位置求和即可。注意空格也算一个字符。

代码：

/*

    Title:HDU.2734

    Date:2016-9-30

    Author:pengwill

    Blog: http://blog.csdn.net/pengwill97

*/

#include<stdio.h>

#include<string.h>

int main()

{

    int sum= 0;

    char ch[300];

    while(gets(ch) && ch[0] != '#'){

        int len = strlen(ch);

        int i;

        sum = 0;

        for(i = 0;i<len;i++){

            if(ch[i] == ' '){

                continue;

            }else{

                sum+=(i+1) * (ch[i] - 64);

            }

        }

        printf("%d\n",sum);

    }

    return 0;

}