计算机学院大学生程序设计竞赛(2015’12)Study Words
Study Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 195 Accepted Submission(s): 66
One day an idea came up to me: I download an article every day, choose the 10 most popular new words to study.
A word's popularity is calculated by the number of its occurrences.
Sometimes two or more words have the same occurrences, and then the word with a smaller lexicographic has a higher popularity.
Each case has two parts.
<oldwords>
...
</oldwords>
<article>
...
</article>
Between <oldwords> and </oldwords> are some old words (no more than 10000) I have already learned, that is, I don't need to learn them any more.
Words between <oldwords> and </oldwords> contain letters ('a'~'z','A'~'Z') only, separated by blank characters (' ','\n' or '\t').
Between <article> and </article> is an article (contains fewer than 1000000 characters).
Only continuous letters ('a'~'z','A'~'Z') make up a word. Thus words like "don't" are regarded as two words "don" and "t”, that's OK.
Treat the uppercase as lowercase, so "Thanks" equals to "thanks". No words will be longer than 100.
As the article is downloaded from the internet, it may contain some Chinese words, which I don't need to study.
If there are fewer than 10 new words, output all of them.
Output a blank line after each case.
2
<oldwords>
how aRe you
</oldwords>
<article>
--How old are you?
--Twenty.
</article>
<oldwords>
google cn huluobo net i
</oldwords>
<article>
文章内容:
I love google,dropbox,firefox very much.
Everyday I open my computer , open firefox , and enjoy surfing on the inter-
net.
But these days it's strange that searching "huluobo" is unavail-
able.
What's wrong with "huluobo"?
</article>
old
twenty firefox
open
s
able
and
but
computer
days
dropbox
enjoy
今天是2015年最后一个星期日,我用了好长好长好长时间做了这道题,测试答案不应该错,可是.........
好伤心、
不想再修改了。。。。就让它一直错吧!
下面的是正确的答案哦!
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
#include<algorithm>
using namespace std;
int T;
char s[100+10];
char r[100+10];
map<string,int>m;
struct dan
{
char s[100+10];
int num;
}d[1000000+10];
int sum;
int tot;
bool cmp(const dan&a,const dan&b)
{
if(a.num==b.num) return strcmp(a.s,b.s)<0;
return a.num>b.num;
} //转小写
void F()
{
for(int i=0;s[i];i++)
if(s[i]>='A'&&s[i]<='Z')
s[i]=s[i]-'A'+'a';
}
void work()
{
int len=strlen(s);
tot=0;
for(int i=0;i<=len;i++)
{
if(s[i]>='a'&&s[i]<='z') r[tot++]=s[i];
else
{
r[tot]='\0';
if(strlen(r)>0) m[r]=-1;
tot=0;
}
}
}
void work2()
{
int len=strlen(s);
tot=0;
for(int i=0;i<=len;i++)
{
if(s[i]>='a'&&s[i]<='z') r[tot++]=s[i];
else
{
r[tot]='\0';
if(strlen(r)>0)
{
if(m[r]!=-1)
{
if(m[r]==0) strcpy(d[sum++].s,r);
m[r]++;
}
}
tot=0;
}
}
}
int main()
{
scanf("%d",&T);
while(T--)
{
int flag=0;
m.clear();
sum=0;
while(1)
{
scanf("%s",s);
if(strcmp(s,"<oldwords>")==0) {flag=1;continue;}
if(strcmp(s,"</oldwords>")==0) {flag=2;continue;}
if(strcmp(s,"<article>")==0) {flag=3;continue;}
if(strcmp(s,"</article>")==0) break;
if(flag==1)
{
F();
work();
}
if(flag==3)
{
F();
work2();
}
}
for(int i=0;i<sum;i++) d[i].num=m[d[i].s];
sort(d,d+sum,cmp);
for(int i=0;i<min(10,sum);i++) printf("%s\n",d[i].s);
printf("\n");
}
return 0;
}
@执念 "@☆但求“❤”安★下次我们做的一定会更好。。。。
为什么这次的题目是英文的。。。。QAQ...
计算机学院大学生程序设计竞赛(2015’12)Study Words的更多相关文章
- hdu 计算机学院大学生程序设计竞赛(2015’11)
搬砖 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submissi ...
- 计算机学院大学生程序设计竞赛(2015’11)1005 ACM组队安排
1005 ACM组队安排 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Pro ...
- 计算机学院大学生程序设计竞赛(2015’12) 1008 Study Words
#include<cstdio> #include<cstring> #include<map> #include<string> #include&l ...
- 计算机学院大学生程序设计竞赛(2015’12)Polygon
Polygon Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Su ...
- 计算机学院大学生程序设计竞赛(2015’12)The Country List
The Country List Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- 计算机学院大学生程序设计竞赛(2015’12) 1009 The Magic Tower
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using ...
- 计算机学院大学生程序设计竞赛(2015’12) 1006 01 Matrix
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> ...
- 计算机学院大学生程序设计竞赛(2015’12) 1003 The collector’s puzzle
#include<cstdio> #include<algorithm> using namespace std; using namespace std; +; int a[ ...
- 计算机学院大学生程序设计竞赛(2015’12) 1004 Happy Value
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include ...
随机推荐
- 【HDOJ6351】Beautiful Now(贪心,搜索)
题意:给定一个数字n,最多可以交换其两个数位k次,求交换后的最大值与最小值,最小值不能有前导0 n,k<=1e9 思路: 当k>=n的位数时只需要无脑排序 k<n时有一个显然的贪心是 ...
- javascript 日期处理类库 moment.js
- 玩转css样式选择器----当父元素有多个子元素时选中第一个
- echarts 图表用例
参考博客:http://blog.csdn.net/verne_feng/article/details/51731653 http://echarts.baidu.com/echarts2/doc/ ...
- jvisualvm远程监控 Visual GC plugin NOT supported for this JVM
1. 找到jdk安装目录. 2. 进入jdk的 bin目录,新建文件jstatd.all.policy. 3.编辑jstatd.all.policy文件,内容如下: 4. 给jstatd.all.po ...
- eslint 在webstorm配置
1.安装nodejs和eslint 2.在 webstorm 的 file - setting搜索eslint,配置eslint路径 3.在项目目录下新建.eslintrc文件 4.配置eslint ...
- luogu P1032 字串变换
题目描述 已知有两个字串 A, B 及一组字串变换的规则(至多6个规则): A1 -> B1 A2 -> B2 规则的含义为:在 A$中的子串 A1 可以变换为 B1.A2 可以变换为 B ...
- CSS 居中 可随着浏览器变大变小而居中
关键代码: 外部DIV使用: text-align:center; 内部DIV使用: margin-left:auto;margin-right:auto 例: <div style=" ...
- Liunx常用命令(备用)
常用指令 ls 显示文件或目录 -l 列出文件详细信息l(list) -a 列出当前目录下所有文件及目录,包括隐藏的a(all) mkdir ...
- 校园网、教育网 如何纯粹访问 IPv6 网站避免收费
我国校园网有可靠的 IPv6 网络环境,速度非常快.稳定,并且大多数高校在网络流量计费时不会限制 IPv6 的流量,也就是免费的.然而访问 IPv4 商业网络时,则会收费,并且连接的可靠性一般.可幸的 ...