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UVA - 409 Excuses, Excuses!

2016-07-05 23:51 399 查看
题目大意:输入一些关键字和几句话,求出现最多个关键字的那句话,重复出现累加。

解题思路:储存关键字和话,去比较就可以了。一开始当成子串去比较,实际上完整单词相同才能算。

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
char word[25][25];
char str[25][100];
char s[25][100];
int num[25];
int count = 0;
int K, E;
int cmp(char *str) {
char t[100];
int l = strlen(str);
int i = 0, j = 0, tot = 0, tag = 0;
for (i = 0; i < l; i++) {
if(!(str[i] >= 'a' && str[i] <= 'z')) {
t[tag] = '\0';
for (int j = 0; j < K; j++) {
if(strcmp(t, word[j]) == 0) tot++;
}
tag = 0;
continue;
}
t[tag] = str[i];
tag++;
}
return tot;
}
int main() {
while (scanf("%d%d", &K, &E) != EOF) {
getchar();
memset (num, 0, sizeof(num));
memset (word, '0', sizeof(word));
memset (str, '0', sizeof(str));
int i, j;
for (i = 0; i < K; i++)
gets(word[i]);
for (i = 0; i < E; i++)
gets(s[i]);
for (i = 0; i < E; i++){
for (j = 0; s[i][j] != '\0'; j++){
if (s[i][j] >= 'A' && s[i][j] <= 'Z')
str[i][j] = s[i][j] - 'A' + 'a';
else str[i][j] = s[i][j];
}
str[i][j] = '\0';
}

for (i = 0; i < E; i++)
num[i] = cmp(str[i]);

int max = 0;
for (i = 0; i < E; i++)
if (num[i] > max)
max = num[i];
printf("Excuse Set #%d\n", ++count);
for (i = 0; i < E; i++)
if (num[i] == max)
puts(s[i]);
printf("\n");
}
return 0;
}
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