2014
02-14

# Quicksum

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

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.

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.

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

46
650
4690
49
75
14
15

#include <iostream>
#include <string>
using namespace std;
int main()
{
char a[2013];
int l;
while(1)
{
gets(a);
if(a[0]=='#')
break;
l=strlen(a);
int i;
int sum=0;
for(i=0;i<l;i++)
switch(a[i])
{
case 'A':sum=sum+(i+1)*1;break;
case 'B':sum=sum+(i+1)*2;break;
case 'C':sum=sum+(i+1)*3;break;
case 'D':sum=sum+(i+1)*4;break;
case 'E':sum=sum+(i+1)*5;break;
case 'F':sum=sum+(i+1)*6;break;
case 'G':sum=sum+(i+1)*7;break;
case 'H':sum=sum+(i+1)*8;break;
case 'I':sum=sum+(i+1)*9;break;
case 'J':sum=sum+(i+1)*10;break;
case 'K':sum=sum+(i+1)*11;break;
case 'L':sum=sum+(i+1)*12;break;
case 'M':sum=sum+(i+1)*13;break;
case 'N':sum=sum+(i+1)*14;break;
case 'O':sum=sum+(i+1)*15;break;
case 'P':sum=sum+(i+1)*16;break;
case 'Q':sum=sum+(i+1)*17;break;
case 'R':sum=sum+(i+1)*18;break;
case 'S':sum=sum+(i+1)*19;break;
case 'T':sum=sum+(i+1)*20;break;
case 'U':sum=sum+(i+1)*21;break;
case 'V':sum=sum+(i+1)*22;break;
case 'W':sum=sum+(i+1)*23;break;
case 'X':sum=sum+(i+1)*24;break;
case 'Y':sum=sum+(i+1)*25;break;
case 'Z':sum=sum+(i+1)*26;break;
case ' ':sum=sum;break;
default:break;
}
cout<<sum<<endl;
sum=0;
}
return 0;
}
#include <iostream>
#include <string.h>
#include <ctype.h>
using namespace std;

int main()
{
char a[255];
int sum;
while(gets(a)&&a[0]!='#')
{
sum=0 ;
for(int j=0;j<strlen(a);j++)
{
if(a[j]!=' ')
sum+=(((int)a[j]-64)*(j+1));
}
cout<<sum<<endl;
}
return 0;
}

1. 有两个重复的话结果是正确的，但解法不够严谨，后面重复的覆盖掉前面的，由于题目数据限制也比较严，所以能提交通过。已更新算法

2. 这道题目虽然简单，但是小编做的很到位，应该会给很多人启发吧！对于面试当中不给开辟额外空间的问题不是绝对的，实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟，今天看到小编这篇文章相见恨晚。