NYOJ 25 A Famous Music Composer
A Famous Music Composer
- 描述
-
Mr. B is a famous music composer. One of his most famous work was his set of preludes. These 24 pieces span the 24 musical keys (there are musically distinct 12 scale notes, and each may use major or minor tonality). The 12 distinct scale notes are:
A A#=Bb B C C#=Db D D#=Eb E F F#=Gb G G#=Ab Five of the notes have two alternate names, as is indicated above with equals sign. Thus, there are 17 possible names of scale notes, but only 12 musically distinct notes. When using one of these as the keynote for a musical key, we can further distinguish between major and minor tonalities. This gives 34 possible keys, of which 24 are musically distinct.In naming his preludes, Mr. B used all the keys except the following 10, which were named instead by their alternate names:Ab minor A# major A# minor C# major Db minor D# major D# minor Gb major Gb minor G# major Write a program that, given the name of a key, give an alternate name if it has one, or report the key name is unique.
- 输入
- Each test case is described by one line having the format "note tonality", where "note" is one of the 17 names for the scale notes given above, and "tonality" is either "major" or "minor" (quotes for clarify).
- 输出
- For each case output the required answer, following the format of the sample.
- 样例输入
-
Ab minor
D# major
G minor - 样例输出
-
Case 1: G# minor
Case 2: Eb major
Case 3: UNIQUE
让我告诉你你WA的原因:
printf("Case %d: ",sign++);
"Case 1: "冒号后面还有个空格。
#include <stdio.h>
#include <string.h>
int main()
{
char s1[],s2[];
int sign = ;
while(scanf("%s%s",s1,s2) != EOF)
{
printf("Case %d: ",sign++);
if(!strcmp(s1,"A#")) printf("%s %s\n","Bb",s2);
else if(!strcmp(s1,"Bb")) printf("%s %s\n","A#",s2);
else if(!strcmp(s1,"C#")) printf("%s %s\n","Db",s2);
else if(!strcmp(s1,"Db")) printf("%s %s\n","C#",s2);
else if(!strcmp(s1,"D#")) printf("%s %s\n","Eb",s2);
else if(!strcmp(s1,"Eb")) printf("%s %s\n","D#",s2);
else if(!strcmp(s1,"F#")) printf("%s %s\n","Gb",s2);
else if(!strcmp(s1,"Gb")) printf("%s %s\n","F#",s2);
else if(!strcmp(s1,"G#")) printf("%s %s\n","Ab",s2);
else if(!strcmp(s1,"Ab")) printf("%s %s\n","G#",s2);
else printf("UNIQUE\n");
}
}
NYOJ 25 A Famous Music Composer的更多相关文章
- 25.A Famous Music Composer
描述 Mr. B is a famous music composer. One of his most famous work was his set of preludes. These 24 p ...
- nyoj 25-A Famous Music Composer(字符串)
25-A Famous Music Composer 内存限制:64MB 时间限制:1000ms Special Judge: No accepted:4 submit:9 题目描述: Mr. B i ...
- 07-语言入门-07-A Famous Music Composer
题目地址: http://blog.csdn.net/sevenmit/article/details/8231994 描述 Mr. B is a famous music composer. On ...
- A Famous Music Composer
描述 Mr. B is a famous music composer. One of his most famous work was his set of preludes. These 24 p ...
- nyoj25-A Famous Music Composer
A Famous Music Composer 时间限制:1000 ms | 内存限制:65535 KB 难度:1 描述 Mr. B is a famous music composer. One ...
- 【南阳OJ分类之语言入门】80题题目+AC代码汇总
小技巧:本文之前由csdn自动生成了一个目录,不必下拉一个一个去找,可通过目录标题直接定位. 本文转载自本人的csdn博客,复制过来的,排版就不弄了,欢迎转载. 声明: 题目部分皆为南阳OJ题目. 代 ...
- 南阳oj水题集合,语言的灵活运用
a+b 输入 输入两个数,a,b 输出 输出a+b的值 样例输入 2 3 样例输出 5 c/c++ #include<iostream> using namespace std; int ...
- nyoj 484-The Famous Clock
484-The Famous Clock 内存限制:64MB 时间限制:1000ms 特判: No 通过数:2 提交数:2 难度:1 题目描述: Mr. B, Mr. G and Mr. M are ...
- NYOJ-79 拦截导弹 AC 分类: NYOJ 2014-01-01 23:25 167人阅读 评论(0) 收藏
#include<stdio.h> int main(){ int num[1000]={0}; int n,m,x,y; scanf("%d",&n); wh ...
随机推荐
- Python 实现 Discuz论坛附件下载权限绕过漏洞
背景:最近压力有些大,想玩点游戏放松下,去Mac论坛下载,发现需要各种权限,于是蛋疼了. 所以,上网查了discuz! x3.1破解,手动替换,发现出现“链接已过期”.所以写了下面程序. 0.将下列代 ...
- Android Shape Divider
安卓框架提供了一种LinearLayout 内部布局元素分割线的实现,建立一个指定长宽的矩形Shape: <?xml version="1.0" encoding=" ...
- Ext修改所有Ajax的timeout
Ext修改所有Ajax的timeout stackoverflow上的解决方案 //需要在初始化viewport时执行 //方法一重写 Ext.Ajax.timeout= 60000; Ext.ove ...
- pouchdb-find( pouchdb查询扩展插件 ,便于查询)
pouchdb-find pouchdb-find 环境搭建 下载lib bower install pouchdb-find 引入js <script src="pouchdb.js ...
- dashDB - Introduction and DB Tools
dashDB - Introduction dashDB is a database that is designed for performance and scale. It offers sea ...
- Java自学手记——Java中的关键字
Java中的一些关键字对于初学者来说有时候会比较混乱,在这里整理一下,顺便梳理一下目前掌握的关键字. 权限修饰符 有四个,权限从大到小是public>protected>defaul(无修 ...
- Struts框架之 执行流程 struts.xml 配置详细
1.执行流程 服务器启动: 1. 加载项目web.xml 2. 创建Struts核心过滤器对象, 执行filter → init() struts-default.xml, 核心功能的初 ...
- 包装类、数组、string类浅析及练习
String s1 = "abc"; String s2 = "abc"; System.out.println(s1==s2); //返回true Strin ...
- HDU 3829 Cat VS Dog / NBUT 1305 Cat VS Dog(二分图最大匹配)
HDU 3829 Cat VS Dog / NBUT 1305 Cat VS Dog(二分图最大匹配) Description The zoo have N cats and M dogs, toda ...
- Android使用ViewPager实现导航菜单
首先设置页面的Fragment布局: public class TabFragment extends ListFragment { @Override public void onViewCreat ...