POJ3094 UVALive3594 HDU2734 ZOJ2812 Quicksum【进制】

Quicksum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16976		Accepted: 11781

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

Source

Mid-Central USA 2006

Regionals 2006 >> North America - Mid-Central USA

问题链接：POJ3094 UVALive3594 HDU2734 ZOJ2812 Quicksum。

问题简述：输入包含若干行，以'#'结束输入。每行输入大写字母开头的，包含大写字母与空格的字符串，字符数<=255个。编写一个程序将字符串转化为一串数字和。转化算法为：对于每行的每一字符，其位权按顺序分别是１、２、３、４、......；每个字符的值是，空格值为0，字母值为1-26，A=1、B=2、C=3、D=4、......、Z=26。如ACF E=1*1+2*3+3*6+4*0+5*5=50。

问题分析：（略）

程序说明：（略）

AC的C语言程序如下：

/* POJ3094 UVALive3594 HDU2734 ZOJ2812 Quicksum */

#include <stdio.h>

int main(void)
{
    int sum, base;
    char c;

    sum = 0;
    base = 0;
    while((c=getchar()) && c != '#') {
        base++ ;
        if('A' <= c && c <= 'Z')
            sum += (c - 'A' + 1) * base;
        else if(c == '
') {
            printf("%d
", sum);
            sum = 0;
            base = 0;
        }
    }

    return 0;
}