def is_A_the_thief(A, B, C, D, E): if (A and not C and E) or (B and C and not E and not D): return 1 else: return 0这个代码用C语言怎么实现

可以使用以下C语言代码来实现上述Python代码的功能：

#include <stdio.h>

int is_A_the_thief(int A, int B, int C, int D, int E) {
    if ((A && !C && E) || (B && C && !E && !D)) {
        return 1;
    } else {
        return 0;
    }
}

int main() {
    int A = 1, B = 0, C = 1, D = 0, E = 1; // 示例输入
    int result = is_A_the_thief(A, B, C, D, E);
    printf("%d\n", result); // 输出结果
    return 0;
}

在C语言中，逻辑运算符的表示方式如下：

- 逻辑与：`&&`
- 逻辑或：`||`
- 逻辑非：`!`

因此，我们只需要将Python代码中的逻辑运算符改为对应的C语言逻辑运算符即可。同时，我们需要在main函数中调用is_A_the_thief函数，并将输入传递给它进行运算，并输出结果。

相关问题

现在需要统计若干段文字(英文)中的单词数量，并且还需统计每个单词出现的次数。 注1：单词之间以空格(1个或多个空格)为间隔。 注2：忽略空行或者空格行。 基本版: 统计时，区分字母大小写，且不删除指定标点符号。 进阶版: 统计前，需要从文字中删除指定标点符号!.,:*?。 注意：所谓的删除，就是用1个空格替换掉相应字符。 统计单词时需要忽略单词的大小写。 输入说明 若干行英文，最后以!!!!!为结束。 输出说明 单词数量 出现次数排名前10的单词（次数按照降序排序，如果次数相同，则按照键值的字母升序排序）及出现次数。 输入样例1 failure is probably the fortification in your pole it is like a peek your wallet as the thief when you are thinking how to spend several hard-won lepta when you are wondering whether new money it has laid background because of you then at the heart of the most lax alert and most low awareness and left it godsend failed !!!!! 输出样例1 46 the=4 it=3 you=3 and=2 are=2 is=2 most=2 of=2 when=2 your=2 输入样例2 Failure is probably The fortification in your pole! It is like a peek your wallet as the thief when You are thinking how to. spend several hard-won lepta. when yoU are? wondering whether new money it has laid background Because of: yOu?, then at the heart of the Tom say: Who is the best? No one dare to say yes. most lax alert and! most low awareness and* left it godsend failed !!!!! 输出样例2 54 the=5 is=3 it=3 you=3 and=2 are=2 most=2 of=2 say=2 to=2

基本版解法：

import re
from collections import defaultdict

# 初始化单词计数器
word_count = defaultdict(int)

# 读入文本
text = ''
while True:
    line = input().strip()
    if line == '!!!!!':
        break
    text += line + ' '

# 统计单词数量
words = re.findall(r'\b\w+\b', text)
for word in words:
    word_count[word] += 1

# 输出单词数量
print(len(words))

# 输出前10个出现次数最多的单词
sorted_words = sorted(word_count.items(), key=lambda x: (-x[1], x[0]))
for i in range(min(10, len(sorted_words))):
    print(sorted_words[i][0], sorted_words[i][1], sep='=')

进阶版解法：

import re
from collections import defaultdict

# 初始化单词计数器
word_count = defaultdict(int)

# 读入文本
text = ''
while True:
    line = input().strip()
    if line == '!!!!!':
        break
    # 删除标点符号
    line = re.sub(r'[.,:*?。]', ' ', line)
    text += line + ' '

# 统计单词数量
words = re.findall(r'\b\w+\b', text.lower())
for word in words:
    word_count[word] += 1

# 输出单词数量
print(len(words))

# 输出前10个出现次数最多的单词
sorted_words = sorted(word_count.items(), key=lambda x: (-x[1], x[0]))
for i in range(min(10, len(sorted_words))):
    print(sorted_words[i][0], sorted_words[i][1], sep='=')

然后输出前10个单词（按字母顺序），如果所有单词不超过10个，则将所有的单词输出。 输入样例 Failure is probably the fortification in your pole It is like a peek your wallet as the thief when you are thinking how to spend several hard-won lepta when you Are wondering whether new money it has laid background Because of you, then at the heart of the most lax alert and most low awareness and left it godsend failed !!!!!

好的，经过处理，输出的结果如下：

Are Because Failure It When a and are as background fortification godsend hard-won has heart how in is it laid lax lepta like low most new of peek pole probably several spend the thief thinking to wallet wondering you your

请注意，输出结果中的单词按照字母顺序排列，并且单词间用空格隔开。另外，输入样例中的标点符号也被认为是单词的一部分。