hdu2262 Quicksum
来源:互联网 发布:如何查看手机是几g网络 编辑:程序博客网 时间:2024/06/15 12:33
Quicksum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2262 Accepted Submission(s): 1418 Problem DescriptionA 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
InputThe 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. OutputFor each packet, output its Quicksum on a separate line in the output.
Sample InputACMMID CENTRALREGIONAL PROGRAMMING CONTESTACNA C MABCBBC#
Sample Output46650469049751415
这里只要注意输入的字符串里面含有空格就好,不能使用cin,要使用gets()函数
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
ACMMID CENTRALREGIONAL PROGRAMMING CONTESTACNA C MABCBBC#
46650469049751415
#include<iostream>using namespace std;int main(){char s[1000];while(gets(s)){if(s[0]=='#')return 0;int i=0,sum=0;while(s[i]!='\0'){if(s[i]>64&&s[i]<91)sum+=(s[i]-64)*(i+1);i++;}cout<<sum<<endl;memset(s,'\0',sizeof(s));}return 0;}
注意每次处理完之后把数组和sum的值清空,这样就ok了
0 0
- hdu2262 Quicksum
- hdu2262
- Quicksum
- Quicksum
- QuickSum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- MySql 分页存储过程
- Ubuntu下配置Tomcat
- iOS中文网址路径转换URLEncode
- 使用BabeLua在cocos2d-x中调试Lua
- 借钱的境界:开价越低 借成的机会反而越小
- hdu2262 Quicksum
- BaseAdapter的ArrayIndexOutOfBoundsException
- Windows下自动连接WiFi 脚本
- Scut游戏服务端免费开源引擎
- 向着爷爷站着的地
- bootmgr解压缩
- C++中定义全局变量
- 升级到cocos2d-x 2.0.2代码差异
- JSTL标签库