NY25 A Famous Music Composer
来源:互联网 发布:ip扫描软件 编辑:程序博客网 时间:2024/05/20 03:42
搞了半天才读明白,原来这个题只是一个简单的替换而已……
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
int j=1;
char s[205];
while(gets(s))
{
if(s[0]=='A'&&s[1]=='#')
{
s[0]='B';
s[1]='b';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='A'&&s[1]=='b')
{
s[0]='G';
s[1]='#';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='B'&&s[1]=='b')
{
s[0]='A';
s[1]='#';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='C'&&s[1]=='#')
{
s[0]='D';
s[1]='b';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='D'&&s[1]=='b')
{
s[0]='C';
s[1]='#';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='D'&&s[1]=='#')
{
s[0]='E';
s[1]='b';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='E'&&s[1]=='b')
{
s[0]='D';
s[1]='#';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='F'&&s[1]=='#')
{
s[0]='G';
s[1]='b';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='G'&&s[1]=='b')
{
s[0]='F';
s[1]='#';
printf("Case %d: %s\n",j++,s);
}
else
if(s[0]=='G'&&s[1]=='#')
{
s[0]='A';
s[1]='b';
printf("Case %d: %s\n",j++,s);
}
else
printf("Case %d: UNIQUE\n",j++);
}
return 0;
}
- NY25 A Famous Music Composer
- A Famous Music Composer
- A famous music composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer
- A Famous Music Composer(map)
- NYOJ - A Famous Music Composer
- 25 A Famous Music Composer
- HDU4245:A Famous Music Composer
- NYOJ25 A Famous Music Composer
- 25 A Famous Music Composer
- HDU4245 A Famous Music Composer
- ACM-A Famous Music Composer
- 京东商城登录逻辑分析,实现程序登录京东商城
- 关于华硕错误0x0000009c蓝屏
- 介绍MFC框架中涉及到的设计模式(二)
- POJ 1845 Sumdiv(快速幂取模+快速分解因式)
- Make array of bools out of Ruby integer bits
- NY25 A Famous Music Composer
- adaboost学习
- CSS自适应布局(左右固定 中间自适应或者右侧固定 左侧自适应)
- 除法运算的注意点
- HDU 1875(最小生成树)
- __repr__与__str__
- POJ2406 KMP前缀周期
- java并发编程(5)--多个线程共享数据和对象的方式
- Deep Learning in NLP(一)