POJ3094 Quicksum
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 18517   Accepted: 12712

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

 
题目分析:
 
     计算只含[大写字母+空格]的字符串的校验和.
     其中每个字母有其对应的特征值,空格=0,A=1,B=2,...,Z=26
     校验和 = 所有 [字符的位置(从1开始) * 字母特征值] 之和
 
#include <memory.h>
#include <iostream>
using namespace std; // 被校验字符串的最大长度
const static int MAX_LEN = ; int getFeatureVal(char c); int main(void) {
char str[MAX_LEN] = { '\0' };
while(true) {
gets(str); // 输入字符串中包含空格,不能用cin接收
int len = strlen(str);
if(strlen(str) <= || str[] == '#') {
break;
} int quicksum = ;
for(int i = ; i < len; i++) {
quicksum += ((i + ) * getFeatureVal(str[i]));
}
cout << quicksum << endl; memset(str, '\0', sizeof(char) * len);
}
return ;
} int getFeatureVal(char c) {
return (c == ' ' ? : (c - 'A' + ));
}

POJ3094 Quicksum的更多相关文章

  1. [字符哈希] POJ 3094 Quicksum

    Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16488   Accepted: 11453 Descri ...

  2. Quicksum -SilverN

    quicksum Given a string of digits, find the minimum number of additions required for the string to e ...

  3. ACM——Quicksum

    Quicksum 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:615            测试通过:256 描述 A chec ...

  4. poj3094

    Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13523   Accepted: 9407 Descrip ...

  5. Quicksum

    Quicksum Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Subm ...

  6. TJU Problem 2520 Quicksum

    注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time L ...

  7. H - Quicksum(1.5.3)

    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit cid=1006#sta ...

  8. HDU.2734 Quicksum

    Quicksum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Subm ...

  9. POJ3094 Sky Code(莫比乌斯反演)

    POJ3094 Sky Code(莫比乌斯反演) Sky Code 题意 给你\(n\le 10^5\)个数,这些数\(\le 10^5\),问这些这些数组成的互不相同的无序四元组(a,b,c,d)使 ...

随机推荐

  1. Linux 添加DNS配置

    Centos7.5 系统,保存退出后自动生效 vi /etc/resolv.conf #阿里云DNS nameserver 223.5.5.5nameserver 223.6.6.6 #百度DNSna ...

  2. Java基础知识——算术操作符、自增自减、关系操作符,你真的了解吗?

    三人行,必有我师焉.欢迎大家加我的微信 yh18482155461,或扫描下方二维码,关注我的微信公众号,共同探讨Java相关技术问题. 温故知新 上一节我们讲到了Java中的赋值操作符,用案例的方式 ...

  3. pycrypto安装各种方法试了,最后这种最快速最方便

    需要安装vs,然后按照如下步骤 这个不是在cmd中执行 而是在vs的命令行中执行

  4. C++ 实验3 类和对象

    Part 2 #ifndef GRAPH_H #define GRAPH_H class Graph { public: Graph(char ch, int n); void draw(); pri ...

  5. cifX使用心得

    一.注意事项: 1.xChannelIORead()第一次读取到数据是空的.2.系统安装cifX驱动之前一定先安装有uio模块和libpciaccess库:建议系统内核为3.X.X版本,测试版本为3. ...

  6. [双系统linux] ----安装完成后无法打开wifi

    在安装了linux 双系统以后发现无法打开wifi和蓝牙. rfkill list all 0:ideapad_wlan: Wireless LANSoft blocked: noHard block ...

  7. Ubuntu16.04 换阿里源

    国内阿里源速度比较快,北京联通下载极快.更新也比较稳定 1.备份 cp /etc/apt/source.list /etc/apt/source.list.bak 2.编辑source文件 sudo ...

  8. FCPX插件54种婚礼调色预设ProWedding Mac

    FCPX插件ProWedding Mac是一款由Pixel Film Studio出品的风格婚礼调色预设fcpx插件,prowedding插件包含54种不同效果的调色预设和4种边缘模糊工具,帮您制作出 ...

  9. 我发起并创立了一个 EPWA 的 开源项目

    EPWA ,  是  Easy PWA  的 意思, PWA 取自于 Google 的 PWA, EPWA   是一个用   C#  Cef  Html  js  css   开发 桌面程序 的 架构 ...

  10. 弹出的 Dialog 里,包含 Form,如何在关闭 Dialog 时,执行 resetFields(对整个表单进行重置,将所有字段值重置为初始值并移除校验结果)

    做法: before-close 事件中,调用 resetFields 取消按钮事件中,调用 resetFields <el-dialog title="弹出窗口" :vis ...