UVA-409 Excuses, Excuses!

本文介绍了一个经典的字符串处理问题 UVA-409 Excuses, Excuses! 的解题思路与实现代码。该题要求统计给定句子中特定单词出现的次数,并找出包含最多关键词的句子。

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UVA-409 Excuses, Excuses!

题目大意:先输入 x 个单词,再输入 y 个句子。统计 y 个句子中哪个含有最多单词,输出含有最多单词的句子。若相等,则一同输出。

Sample Input

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I ate my dog, my canary, and my homework … note outdated keywords?

Sample Output

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!

解题思路:输入,注意将大写字母转化为小写字母,将单词逐个比较,累计求最大的,输出。

//UVA-409 Excuses, Excuses!
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n,m,value = 1;
char word[100][100];
char str1[100][100];
char str[100][100];
int numb[100];
int num(char str[100]) {
    int nn = strlen(str);
    int j = 0;
    int number = 0;
    char s[100];
    for (int i = 0; i < nn; i++) {
        if (!(str[i] >= 'a' && str[i] <= 'z' || str[i] >='A' && str[i] <= 'Z')) {
            s[j] = '\0';
            for (int j = 0; j < n; j++) {
                if (strcmp(s,word[j]) == 0)
                    number++;
            }
            j = 0;
            continue;
        }
        s[j] = str[i];
        j++;
    }
    return number;
}
int main () {
    while (scanf ("%d %d",&n,&m) != EOF) {
        getchar();
        for (int i = 0; i < n; i++) {
            gets(word[i]);
            int ss =strlen (word[i]);
            for (int j = 0; j < ss; j++) {
                if (word [i][j] >= 'A' && word [i][j] <= 'Z')
                    word[i][j] = word[i][j] + 'a' - 'A';
            }
        }
        int i,j,k;
        for (i = 0; i < m; i++) {
            gets(str1[i]);
            int ll = strlen(str1[i]);
            for (j = 0; j < ll; j++)
                str[i][j] = str1[i][j];
            str[i][j] = '\0';
            for (k = 0; k < ll; k++)
                if (str[i][k] >= 'A' && str[i][k] <= 'Z')
                    str[i][k] = str[i][k] + 'a' - 'A';
            numb[i] = num(str[i]);
        }
        printf("Excuse Set #%d\n",value++);
        int max = 0, h;
        for ( h = 0; h < m; h++) {
            if (numb[h] > max)
                max = numb[h];
        }
        for ( h = 0; h < m; h++) {
            if (numb[h] == max) {
                puts(str1[h]);
            }
        }
        printf("\n");
    }
    return 0;
}
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