UVA - 409 Excuses, Excuses!

 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

Input to your program will consist of multiple sets of data.

• Line 1 of each set will contain exactly two integers. The first number ( ) 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.
• Lines 2 through K+1 each contain exactly one keyword.
• Lines K+2 through K+1+E each contain exactly one excuse.
• All keywords in the keyword list will contain only contiguous lower case alphabetic characters of lengthL ( ) and will occupy columns 1 throughL in the input line.
• All excuses can contain any upper or lower case alphanumeric character, a space, or any of the following punctuation marks [SPMamp".,!?&] not including the square brackets and will not exceed 70 characters in length.
• Excuses will contain at least 1 non-space character.

Output

For each input set, you are to print the worst excuse(s) from the list.

• The worst excuse(s) is/are defined as the excuse(s) which contains the largest number of incidences of keywords.
• If a keyword occurs more than once in an excuse, each occurrance is considered a separate incidence.
• A keyword ``occurs" in an excuse if and only if it exists in the string in contiguous form and is delimited by the beginning or end of the line or any non-alphabetic character or a space.

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.

Sample Input

5 3dogatehomeworkcanarydiedMy 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 5superhighwaycrazythermonuclearbedroomwarbuildingI 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?

Sample Output

Excuse Set #1Can you believe my dog died after eating my canary... AND MY HOMEWORK?Excuse Set #2I am having a superhighway built in my bedroom.There was a thermonuclear war!

题意：

给出几个单词和几个句子，找出哪一句里面含有的单词最多，并输出来（如果里面有相同的数目一起输出来）

下面是我之前的错误，经过修改之后的代码。

#include<iostream>#include<stdio.h>#include<ctype.h>#include<string.h>#define N 50   //错误一 : 数组开太小using  namespace std;int m, n, k;char word[N][N];   int num(char *a) {int number = 0, len = strlen(a), j = 0;char s[N];for (int i = 0; i <= len ; i++) {   //错误二 : i 要小于等于 len, 否则当 len = 1 时会出错.if (!(a[i] >= 'A' && a[i] <= 'Z' || a[i] >= 'a' && a[i] <= 'z')) {s[j] = '\0';for (k = 0; k < n; k++)if (strcmp(s, word[k]) == 0)number++;j = 0;continue;}s[j] = a[i];j++;}return number;}int main(){char c[N][N * N], d[N][N * N];int len1, n1 = 0, numa[N], j;while (scanf("%d%d", &n, &m) != EOF) {getchar();n1++;for (int i = 0; i < n; i++) {gets(word[i]);int len2 = strlen(word[i]);for (int j = 0; j < len2; j++)word[i][j] = tolower(word[i][j]);}for (int i = 0; i < m; i++) {gets(c[i]);strcpy(d[i], c[i]);len1 = strlen(c[i]);for (int j = 0; j < len1; j++)d[i][j] = tolower(d[i][j]);numa[i] = num(d[i]);}cout<< "Excuse Set #"<< n1<< endl;int max = 0;for (int i = 0; i < m; i++)if (numa[i] >= max)max = numa[i];for (int i = 0; i < m; i++)  //错误三 : i 是小于 m，不是小于 nif (max == numa[i])puts(c[i]);printf("\n");     //错误四 : 每个输出样例的后面都要输出个空行}return 0;}