poj3094-Quicksum

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本文深入解析了Quicksum算法，一种用于检测数据包错误的检查和算法。通过将每个字符的位置与其在字母表中的位置相乘并求和，Quicksum能够有效识别数据传输中的错误。文章提供了详细的计算示例和实现代码。

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

650

4690

题意：输入一串小于255的字符串，写一个算法，使其可以输出一个指定的数值，数值等于字符所在的位置*该字符代表的数，‘A’为1，‘B’为2...‘Z’为26，以‘#’结束输出，字符串只有大写字母和空格，空格可以是连续的。

1.确定输入格式，空格不代表结束以gets(a)输入字符串

2.当a[0]为‘#’时结束while循环

3.主要代码，以n记录当前字符的位置，sum记录当前字符的和

3.当a[i]为空格时，n++，sum不变，当a[i]为字符时，sum+=n*（a[i]-'A'+1）

4.当字符串结束后，输出sum的值

5.下次循环开始时，重置sum和n的值

其代码如下：

#include<cstdio>
using namespace std;
int main()
{
    char a[300];
    int n, sum;
    while(gets(a))
    {
        sum = 0;
        n = 1;
        if(a[0]=='#')
            break;
        for(int i = 0; a[i] != '\0'; i++)
        {
            if(a[i] == ' ')
            {
                n++;
                continue;
            }
            if(a[i] >='A' && a[i] <= 'Z')
            {
                sum += n*(a[i] - 'A'+1);
                n++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}