Quicksum(字符串基础)
来源:互联网 发布:微信服务号 域名 端口 编辑:程序博客网 时间:2024/06/06 04:11
http://poj.org/problem?id=3094
Quicksum
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 11831
Accepted: 8151
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
Mid-Central USA 2006
解析:水题:值得注意的一点就是完整的输入一行带空格的字符串
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
using namespace std;
const int maxn=300;
char st[maxn];
int main()
{
int sum,i,j;
int a[30];
int len;
for(i=0;i<len;i++)
a[i]=i+1;
for(;;)
{scanf("\n");//完整的输入一行带空格的字符串这一步必不可少
cin.getline(st,maxn);
if(st[0]=='#')
break;
len=strlen(st);
sum=0;
for(i=0;i<len;i++)
{
if(st[i]!=' ')
sum+=(i+1)*a[st[i]-'A'];
}
printf("%d\n",sum);
}
system("pause");
return 0;
}
- Quicksum(字符串基础)
- hdoj 2734 Quicksum(字符串)
- HDU 2734 Quicksum【水 字符串】
- hdu2734 Quicksum (水)
- Quicksum(poj3094,水题)
- hdu 2734(Quicksum)
- Quicksum
- Quicksum
- QuickSum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- Quicksum
- 直播电视HLS协议分析及实现2---协议实例分析
- 正则表达式30分钟入门教程(一)
- java-web-书籍推荐
- poj1163——The Triangle
- Release版本下SetItem的错误
- Quicksum(字符串基础)
- HEVC中的SAO 5 saoComponentParamDist
- [IOS 按钮]--“相机闪光灯”控制按钮样式 DDExpandableButton
- Jquery简单实现图书导航
- android开发(22)使用正则表达式 。从一个字符串中找出数字,多次匹配。
- linux tar加压、解压命令
- 小型英汉词典(黑白界面下的)使用了二叉搜索树C++
- 第六周项目四 有错误,关于指针引用那一块
- shell排序