【PAT甲级】1100 Mars Numbers (20 分)
题意:
输入一个正整数N(<100),接着输入N组数据每组包括一行字符串,将其翻译为另一个星球的数字。
AAAAAccepted code:
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
char a[][]={"","tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};
char b[][]={"tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec"};
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin>>n;
cin.ignore();
for(int q=;q<=n;++q){
string s;
getline(cin,s);
if(s[]>=''&&s[]<=''){
int temp=;
for(int j=;j<s.size();++j){
temp*=;
temp+=s[j]-'';
}
int x=temp/;
int y=temp%;
if(x&&y)
cout<<a[x]<<" "<<b[y]<<"\n";
else if(x)
cout<<a[x]<<"\n";
else
cout<<b[y]<<"\n";
}
else{
if(s[]==' '&&s[]!=){
char x[]={},y[]={};
for(int i=;i<;++i)
x[i]=s[i];
for(int i=;i<;++i)
y[i-]=s[i];
int pos=;
for(int i=;i<=;++i){
if(strcmp(a[i],x)==){
pos=i;
break;
}
}
int pos2=;
for(int i=;i<=;++i){
if(strcmp(b[i],y)==){
pos2=i;
break;
}
}
int temp=pos*+pos2;
cout<<temp<<"\n";
}
else if(s[]!=&&s[]==)
cout<<"0\n";
else{
char x[]={};
for(int i=;i<;++i)
x[i]=s[i];
int pos=;
for(int i=;i<=;++i)
if(strcmp(a[i],x)==){
pos=i;
break;
}
int pos2=;
for(int i=;i<=;++i)
if(strcmp(b[i],x)==){
pos2=i;
break;
}
int temp=pos*+pos2;
cout<<temp<<"\n";
}
}
}
return ;
}
【PAT甲级】1100 Mars Numbers (20 分)的更多相关文章
- PAT甲级——1100 Mars Numbers (字符串操作、进制转换)
本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90678474 1100 Mars Numbers (20 分) ...
- PAT 甲级 1041 Be Unique (20 分)(简单,一遍过)
1041 Be Unique (20 分) Being unique is so important to people on Mars that even their lottery is de ...
- PAT 甲级 1035 Password (20 分)
1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...
- PAT 甲级 1073 Scientific Notation (20 分) (根据科学计数法写出数)
1073 Scientific Notation (20 分) Scientific notation is the way that scientists easily handle very ...
- PAT 甲级 1046 Shortest Distance (20 分)(前缀和,想了一会儿)
1046 Shortest Distance (20 分) The task is really simple: given N exits on a highway which forms a ...
- PAT 甲级 1042 Shuffling Machine (20 分)(简单题)
1042 Shuffling Machine (20 分) Shuffling is a procedure used to randomize a deck of playing cards. ...
- PAT甲级——1152.Google Recruitment (20分)
1152 Google Recruitment (20分) In July 2004, Google posted on a giant billboard along Highway 101 in ...
- PAT甲级 1120. Friend Numbers (20)
1120. Friend Numbers (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Two in ...
- PAT 甲级 1050 String Subtraction (20 分) (简单送分,getline(cin,s)的使用)
1050 String Subtraction (20 分) Given two strings S1 and S2, S=S1−S2 is defined to be t ...
随机推荐
- [LOJ144] DFS 序 1 - DFS序 线段树
为熟悉机房键盘而划水 #include <bits/stdc++.h> using namespace std; #define int long long const int N = 4 ...
- html+css 文本折叠
先看效果: 收缩状态 展开状态 源代码: <!doctype html> <html lang="zh"> <head> <meta ch ...
- CentOS7下使用C/C++连接MariaDB/MySQL
前言 连接数据库通常在Java中使用比较多,但是C/C++在Linux下操作数据库也是比较重要的,很多时候都能用得到,在网上查了很多教程,大多写的有些问题,通过自己摸索,终于成功的连接了MariaDB ...
- 这是一篇通过open live writer发布的博文
这两天零零总总的尝试了两三款写博客的软件,总感觉不怎么上手,最后还是使用博客园官方推荐的工具写博吧,简单方便,目前的功能基本都有,尤其是粘贴图片特别方便,回想之前的几篇博文,真是一种煎熬哈哈(对于我这 ...
- 7.log4j
Log4j:日志工厂的一部分(使用起来比较麻烦) 1.要想使用外部类,得先导包 pom.xml <dependency> <groupId>log4j</groupId& ...
- 题解【POJ1160】Post Office
[POJ1160]Post Office Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22362 Accepted: 1208 ...
- Unity小知识点大全(一)
原帖地址:https://zhuanlan.zhihu.com/p/54763553 1. 高亮选择 在Scene面板右上角的Gizmo下拉列表中,可以通过设置Selection Outline选项决 ...
- 【转载】Java泛型(一)
转自:http://www.cnblogs.com/lzq198754/p/5780426.html 1.为什么需要泛型 泛型在Java中有很重要的地位,网上很多文章罗列各种理论,不便于理解,本篇将立 ...
- MongoDB高可用集群配置方案
原文链接:https://www.jianshu.com/p/e7e70ca7c7e5 高可用性即HA(High Availability)指的是通过尽量缩短因日常维护操作(计划)和突发的系统崩溃(非 ...
- 普及C组第四题(8.9)
2298. [noip普及组2T4]异或 (File IO): input:gcdxor.in output:gcdxor.out 题目描述 SarvaTathagata是个神仙,一天他在研究数论时, ...