Description
| Excuses, Excuses! |
Judge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time required listening to goofy excuses, Judge Ito has asked that you write a program that will search for a list of keywords in a list of excuses identifying lame excuses. Keywords can be matched in an excuse regardless of case.
Input to your program will consist of multiple sets of data.
) defines the number of keywords to be used in the
search. The second number ( 
) defines the number of excuses in the set to be searched.
) and
will occupy columns 1 through L in the input line.SPMamp".,!?&] not including the square brackets and will not exceed 70 characters in length.For each input set, you are to print the worst excuse(s) from the list.
For each set of input, you are to print a single line with the number of the set immediately after the string ``Excuse Set #". (See the Sample Output). The following line(s) is/are to contain the worst excuse(s) one per line exactly as read in. If there is more than one worst excuse, you may print them in any order.
After each set of output, you should print a blank line.
5 3 dog ate homework canary died My dog ate my homework. Can you believe my dog died after eating my canary... AND MY HOMEWORK? This excuse is so good that it contain 0 keywords. 6 5 superhighway crazy thermonuclear bedroom war building I am having a superhighway built in my bedroom. I am actually crazy. 1234567890.....,,,,,0987654321?????!!!!!! There was a thermonuclear war! I ate my dog, my canary, and my homework ... note outdated keywords?
Excuse Set #1 Can you believe my dog died after eating my canary... AND MY HOMEWORK? Excuse Set #2 I am having a superhighway built in my bedroom. There was a thermonuclear war!
题意：给两个数m，n。即m个关键字，n句借口，哪个借口包含的的关键字最多，就输出当前的借口
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
using namespace std;
char key[20][20];
char ex[20][110];
int cnt[110];
int n,m;
int find(char *str)
{
char a[110];
int i,j=0;
int count = 0;
for(i=0;str[i]!='\0';i++)
{
if(str[i]>='A'&&str[i]<='Z'||str[i]>='a'&&str[i]<='z')//关键字都是小写，所以把借口里面的大写字母全部转换成小写字母
{
if(str[i]>='A'&&str[i]<='Z')
{
a[j]=str[i]+32;
}
else
a[j]=str[i];
j++;
}
else//因为语句中包含其他的字符，所以当是其他字符的时候
{
a[j] = '\0';//意味着重新换成另一个单词
for(j = 0; j < m; j++)
{
if(strcmp(a,key[j]) == 0)
{
count++;
}
}
j = 0;//拿出下一个单词；
}
}
return count;
}
int main()
{
int i;
int t=1;
int maxx;
while(~scanf("%d %d",&m,&n))
{
getchar();
maxx=0;
memset(cnt,0,sizeof(cnt));
for(i=0; i<m; i++)
gets(key[i]);
for(i=0; i<n; i++)
gets(ex[i]);
for(i=0; i<n; i++)
cnt[i]=find(ex[i]);
for(i=0; i<n; i++)
maxx=max(maxx,cnt[i]);
printf("Excuse Set #%d\n",t++);
for(i=0; i<n; i++)
{
if(maxx==cnt[i])
{
puts(ex[i]);
}
}
printf("\n");
}
return 0;
}
原文：http://blog.csdn.net/u013486414/article/details/42213029