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  • POJ3094Quicksum

    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

    一般思路:
    1、开数组,浪费。
    2、不开数组:
    View Code
     1 #include <iostream>
     2 #include <stdio.h>
     3 
     4 using namespace std;
     5 
     6 int main()
     7 {
     8     int i, sum;
     9     char x;
    10     x=getchar();
    11     sum=0;
    12     i=1;
    13     while(x!='#'){
    14         if(x==' '){
    15             i++;
    16         }else{
    17             sum+=(x-'A'+1)*(i++);
    18         }
    19         x=getchar();
    20 
    21         if(x=='\n'){
    22             cout<<sum<<endl;
    23             sum=0;
    24             x=getchar();
    25             i=1;
    26         }
    27     }
    28     return 0;
    29 }

    3、缩短代码

    View Code
     1 #include<stdio.h>
     2 
     3 int main()
     4 {
     5     char c;
     6     while((c = getchar()) != '#')
     7     {
     8         int i = 1;
     9         long int sum = 0;
    10         do
    11         {
    12             sum += i * (c == ' ' ? 0 : c - 'A' + 1);
    13             i ++;
    14         }while((c = getchar()) != 10);
    15         printf("%ld\n", sum);
    16     }
    17     return 0;
    18 }
    ==================================================

    作者: Panderen

    博客: http://panderen.cnblogs.com

    签名: 机会总是为有准备的人而准备的!

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  • 原文地址:https://www.cnblogs.com/panderen/p/2443185.html
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