USACO_1.1_Your_Ride_Is_Here_(字符串+水题)
描述
http://train.usaco.org/usacoprob2?a=y0SKxY0Kc2q&S=ride
给出两个由大写字母组成,长度不大于$6$的字符串.
将字符串中的各字母的字典序数相乘,最后对$47$取模,比较两个字符串的结果是否相同.
Your Ride Is Here
It is a well-known fact that behind every good comet is a UFO. These UFOs often come to collect loyal supporters from here on Earth. Unfortunately, they only have room to pick up one group of followers on each trip. They do, however, let the groups know ahead of time which will be picked up for each comet by a clever scheme: they pick a name for the comet which, along with the name of the group, can be used to determine if it is a particular group's turn to go (who do you think names the comets?). The details of the matching scheme are given below; your job is to write a program which takes the names of a group and a comet and then determines whether the group should go with the UFO behind that comet.
Both the name of the group and the name of the comet are converted into a number in the following manner: the final number is just the product of all the letters in the name, where "A" is 1 and "Z" is 26. For instance, the group "USACO" would be 21 * 19 * 1 * 3 * 15 = 17955. If the group's number mod 47 is the same as the comet's number mod 47, then you need to tell the group to get ready! (Remember that "a mod b" is the remainder left over after dividing a by b; 34 mod 10 is 4.)
Write a program which reads in the name of the comet and the name of the group and figures out whether according to the above scheme the names are a match, printing "GO" if they match and "STAY" if not. The names of the groups and the comets will be a string of capital letters with no spaces or punctuation, up to 6 characters long.
Examples:
| Input | Output |
COMETQ |
GO |
ABSTAR |
STAY |
PROGRAM NAME: ride
This means that you fill in your header with:
PROG: ride
WARNING: You must have 'ride' in this field or the
wrong test data (or no test data) will be used.
INPUT FORMAT
| Line 1: | An upper case character string of length 1..6 that is the name of the comet. |
| Line 2: | An upper case character string of length 1..6 that is the name of the group. |
NOTE: The input file has a newline at the end of each line
but does not have a "return". Sometimes, programmers code for
the Windows paradigm of "return" followed by "newline"; don't do
that! Use simple input routines like "readln" (for Pascal) and,
for C/C++, "fscanf" and "fid>>string".
NOTE 2: Because of the extra characters, be sure to leave
enough room for a 'newline' (also notated as '\n') and an end of
string character ('\0') if your language uses it (as C and C++ do).
This means you need eight characters of room instead of six.
SAMPLE INPUT (file ride.in)
COMETQ
HVNGAT
OUTPUT FORMAT
A single line containing either the word "GO" or the word "STAY".
SAMPLE OUTPUT (file ride.out)
GO
OUTPUT EXPLANATION
Converting the letters to numbers:
| C | O | M | E | T | Q | |
| 3 | 15 | 13 | 5 | 20 | 17 | |
| H | V | N | G | A | T | |
| 8 | 22 | 14 | 7 | 1 | 20 |
then calculate the product mod 47:
3 * 15 * 13 * 5 * 20 * 17 = 994500 mod 47 = 27
8 * 22 * 14 * 7 * 1 * 20 = 344960 mod 47 = 27
Because both products evaluate to 27 (when modded by 47), the mission is 'GO'.
分析
没啥好分析的..
/*
TASK:ride
LANG:C++
*/ #include <bits/stdc++.h>
using namespace std; const int mod=, maxn=;
int x,y;
char a[maxn],b[maxn]; int main(){
freopen("ride.in","r",stdin);
freopen("ride.out","w",stdout);
scanf("%s%s",a,b);
x=y=;
for(int i=;i<maxn;i++){
if(a[i]) x=(x*(a[i]-'A'+))%mod;
if(b[i]) y=(y*(b[i]-'A'+))%mod;
}
x==y ? puts("GO") : puts("STAY");
return ;
}
USACO_1.1_Your_Ride_Is_Here_(字符串+水题)的更多相关文章
- 1222: FJ的字符串 [水题]
1222: FJ的字符串 [水题] 时间限制: 1 Sec 内存限制: 128 MB 提交: 92 解决: 20 统计 题目描述 FJ在沙盘上写了这样一些字符串: A1 = “A” A2 = ...
- 1001 字符串“水”题(二进制,map,哈希)
1001: 字符串“水”题 时间限制: 1 Sec 内存限制: 128 MB提交: 210 解决: 39[提交][状态][讨论版] 题目描述 给出一个长度为 n 的字符串(1<=n<= ...
- 第十一届“蓝狐网络杯”湖南省大学生计算机程序设计竞赛 B - 大还是小? 字符串水题
B - 大还是小? Time Limit:5000MS Memory Limit:65535KB 64bit IO Format: Description 输入两个实数,判断第一个数大 ...
- HDU ACM 1073 Online Judge ->字符串水题
分析:水题. #include<iostream> using namespace std; #define N 5050 char a[N],b[N],tmp[N]; void Read ...
- HDU4891_The Great Pan_字符串水题
2014多校第五题,当时题面上的10^5写成105,我们大家都wa了几发,改正后我和一血就差几秒…不能忍 题目:http://acm.hdu.edu.cn/showproblem.php?pid=48 ...
- Codeforces Round #309 (Div. 2) B. Ohana Cleans Up 字符串水题
B. Ohana Cleans Up Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/554/pr ...
- Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks 字符串水题
A. Kyoya and Photobooks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- uva 10252 - Common Permutation 字符串水题
题意:給定兩個小寫的字串a與b,請印出皆出現在兩字串中的字母,出現的字母由a~z的順序印出,若同字母出現不只一次,請重複印出但不能超過任一字串中出現的次數.(from Ruby兔) 很水,直接比较输出 ...
- hdu1106 字符串水题strtok()&&strchr()&&sscanf()+atoi()使用
字符串的题目 用库函数往往能大大简化代码量 以hdu1106为例 函数介绍 strtok() 原型: char *strtok(char s[], const char *delim); 功能: 分解 ...
随机推荐
- svn 服务器搭建 (Linux)
1.svn目前在程序开发工程汇总应用非常普遍,所以学习svn的环境搭建还是很有必要的 2.本次安装的服务是Subversion(svn)针对的环境是Linux,Subversion(SVN) 是一个开 ...
- Django笔记 —— 模型高级进阶
最近在学习Django,打算玩玩网页后台方面的东西,因为一直很好奇但却没怎么接触过.Django对我来说是一个全新的内容,思路想来也是全新的,或许并不能写得很明白,所以大家就凑合着看吧- 本篇笔记(其 ...
- ios下 active 演示激活
document.body.addEventListener('touchstart', function () { });
- 基于Python的接口自动化-01
为什么要做接口测试 当前互联网产品迭代速度越来越快,由之前的2-3个月到个把月,再到班车制,甚至更短,每次发版之前都需要对所有功能进行回归测试,在人力资源有限的情况下,做自动化测试很有必要.由于UI更 ...
- 13.0 Excel表格写入
Excel表格写入 安装 xlutils 和 xlwt Excel写入输入 分两种方式: 第一种是向一张新表之中写入..这种不多说,我几乎没怎么用,直接贴代码 import xlwt Excel_na ...
- Thymeleaf 使用时的标签
1 . onclick事件 <a th:onclick="'javascript:more()'" ></a> 2.引入CSS样式 <link t ...
- 面向对象 公有私有 property classmethod staticmethod
接口类(抽象类)--------就是一种规范 面向对象的私有与公有 对于每一个类的成员而言都有两种形式: 公有成员,在任何地方都能访问 私有成员,只有在类的内部才能方法 私有成员和公有成员的访问限制不 ...
- 孤荷凌寒自学python第六十八天学习并实践beautifulsoup模块1
孤荷凌寒自学python第六十八天学习并实践beautifulsoup模块1 (完整学习过程屏幕记录视频地址在文末) 感觉用requests获取到网页的html源代码后,更重要的工作其实是分析得到的内 ...
- mysql 查询表的字段数目
select column_name from information_schema.`COLUMNS` where TABLE_NAME ='tcm_head'
- 【转】GOOGLE-PROTOBUF与FLATBUFFERS数据的序列化和反序列化
转载自[黑米GameDev街区] 原文链接: http://www.himigame.com/unity3d-game/1607.html 关于Protobuf 通过本文的转载和分享的相关链接,足够 ...