Quicksum



题目描述

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

输入

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.

输出

For each packet, output its Quicksum on a separate line in the output.

样例输入

ACM
MID CENTRAL
REGIONAL PROGRAMMING CONTEST
ACN
A C M
ABC
BBC
#

样例输出

46
650
4690
49
75
14
15

提示:

题目大意是，输入字符串（输入有多行，每行有一个或多个单词（单词仅包含大写字母和空格）），直到输入#号输入结束。对于每一行单词，输出它们的值，每个值一行。（＃不用输出）。计算出一个  “特殊的值=字符位置（从第一个开始）*该字符在字母表里的位置+字符位置（第二个）*该字符在字母表的位置+……”（空格视为0）

题解：

#include<iostream>
#include<string.h>
#include<cstdio>//gets的头文件(c++)or stdio.h(c) 
using namespace std;
int main()
{
	char date[255];
	int sum=0,t;
	while(gets(date)&&date!=NULL)
	{
		if(date[0]=='#')
		break;
		for(int i=0;i<strlen(date);i++)
		{
			if(date[i]==' ')
			t=0;
			else
			t=(date[i]-'@');//判断于字母表中位置 
			sum+=(i+1)*t;
		}
		cout<<sum<<endl;
		sum=0;
	}
}