The Country List

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2464    Accepted Submission(s): 576

Problem Description
As the 2010 World Expo hosted by Shanghai is coming, CC is very honorable to be a volunteer of such an international pageant. His job is to guide the foreign visitors. Although he has a strong desire to be an excellent volunteer, the lack of English makes him
annoyed for a long time. 

Some countries’ names look so similar that he can’t distinguish them. Such as: Albania and Algeria. If two countries’ names have the same length and there are more than 2 same letters in the same position of each word, CC cannot distinguish them. For example:
Albania and AlgerIa have the same length 7, and their first, second, sixth and seventh letters are same. So CC can’t distinguish them.

Now he has received a name list of countries, please tell him how many words he cannot distinguish. Note that comparisons between letters are case-insensitive.
 
Input
There are multiple test cases.

Each case begins with an integer n (0 < n < 100) indicating the number of countries in the list.

The next n lines each contain a country’s name consisted by ‘a’ ~ ‘z’ or ‘A’ ~ ‘Z’.

Length of each word will not exceed 20.

You can assume that no name will show up twice in the list.
 
Output
For each case, output the number of hard names in CC’s list.
 
Sample Input
3
Denmark
GERMANY
China
4
Aaaa
aBaa
cBaa
cBad
 
Sample Output
2
4
 

总是望着曾经的空间发呆,那些说好不分开的朋友不在了,转身,陌路。 熟悉的,安静了, 安静的,离开了, 离开的,陌生了, 陌生的,消失了, 消失的,陌路了。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int da[105]= {0};
void bijiao(int n,char c[105][30])
{
int i,j,t=0,q;
for(i=0; i<n; i++)
for(j=0; j<n; j++)
{
t=0;
if(strcmp(c[i],c[j])==0)continue;
if(strlen(c[i])==strlen(c[j]))
{
for(q=0; q<(int)strlen(c[i]); q++)
{
if(c[i][q]==c[j][q])t++;
}
if(t>2)
{
da[i]=1;
da[j]=1;
}
}
}
}
void xiaoxie(char c[30])
{
int n=strlen(c),i;
for(i=0; i<n; i++)
{
if(c[i]>='A'&&c[i]<='Z')c[i]+=32;
}
}
int main()
{
int n,i,j;
char c[105][30];
while(~scanf("%d",&n)&&n)
{
j=0;
getchar();
for(i=0; i<n; i++)da[i]=0;
for(i=0; i<n; i++)
{
gets(c[i]);
xiaoxie(c[i]);
}
bijiao(n,c);
for(i=0; i<n; i++)j+=da[i];
printf("%d\n",j);
}
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)The Country List的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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) Tota ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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> ...

  8. 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[ ...

  9. 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

  1. jeasyui datagrid 使用记

    1. 一开始想用$('#dg').datagrid('getChanges')来保存整个table修改的行, 结果发现当前编辑的行不算,要失去焦点才有记录 2. 然后改用       $('#dg') ...

  2. 推荐几个sql server牛人的博客

    Aaron Bertrand http://sqlblog.com/blogs/aaron_bertrand/ Brent Ozar www.brentozar.com/ Buck Woody htt ...

  3. tableview在第一次显示时会自动relodata

    tableview在第一次显示时会自动加载数据

  4. Leetcode: Strobogrammatic Number III

    A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside ...

  5. SQL 数据库基础

    SQL:Structured Quety Language SQL SERVER是一个以客户/服务器(c/s)模式访问.使用Transact-SQL语言的关系型数据库管理子系统(RDBMS) DBMS ...

  6. 两条直线(蓝桥杯)二分枚举+RMQ

    算法提高 两条直线   时间限制:1.0s   内存限制:256.0MB        问题描述 给定平面上n个点. 求两条直线,这两条直线互相垂直,而且它们与x轴的夹角为45度,并且n个点中离这两条 ...

  7. 某个点到其他点的曼哈顿距离之和最小(HDU4311)

    Meeting point-1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  8. ACRush 楼天成回忆录

    楼教主回忆录: 利用假期空闲之时,将这几年 GCJ , ACM , TopCoder 参加的一些重要比赛作个回顾.首先是 GCJ2006 的回忆. Google Code Jam 2006 一波三折: ...

  9. SQL事物

    事务:保障流程的完整执行就像银行取钱,先在你账上扣钱,然后存入别人的账上:但是从你账上扣完钱了,突然网断了,对方没有收到钱,那么此时你的钱也没了,别人的钱也没加上,为了防止此类情况的出现,事务. be ...

  10. mysql之消息队列

    消息队列:在消息的传输过程中保存消息的容器. 消息队列管理器在将消息从它的源中继到它的目标时充当中间人.队列的主要目的是提供路由并保证消息的传递:如果发送消息时接收者不可用,消息队列会保留消息,直到可 ...