L1-064 估值一亿的AI核心代码
题目 L1-064 估值一亿的AI核心代码
思路分析
代码实现
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
using ll = long long;
using ull = unsigned long long;
using PII = pair<int, int>;
using Pll = pair<ll, ll>;
int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
const int inf = 0x3f3f3f3f;
// 去除多余空格，标点前空格也要删除
string normalize_space(const string& input) {
stringstream ss(input);
string word, result;
while (ss >> word) {
if (!result.empty()) result += " ";
result += word;
}
// 删除标点前的空格
string res2;
for (size_t i = 0; i < result.size(); ++i) {
if (i > 0 && ispunct(result[i]) && result[i - 1] == ' ') {
res2.pop_back(); // 删除标点前的空格
}
res2 += result[i];
}
return res2;
}
// 判断某位置是否为独立的单词
bool is_word_boundary(const string& s, int start, int len) {
bool left = (start == 0 || !isalnum(s[start - 1]));
bool right = (start + len >= s.size() || !isalnum(s[start + len]));
return left && right;
}
// 将所有大写字母转小写，除了大写的I
string to_lower_custom(const string& input) {
string res;
for (char c : input) {
if (c == 'I') {
res += 'I';
} else if (isupper(c)) {
res += tolower(c);
} else {
res += c;
}
}
return res;
}
// 替换函数（匹配独立单词）
void replace_word(string& str, const string& from, const string& to) {
size_t pos = 0;
while ((pos = str.find(from, pos)) != string::npos) {
if (is_word_boundary(str, pos, from.length())) {
str.replace(pos, from.length(), to);
pos += to.length();
} else {
pos += from.length();
}
}
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int N;
cin >> N;
cin.ignore();
for (int i = 0; i < N; ++i) {
string line;
getline(cin, line);
cout << line << endl;
// 1. 消除多余空格
string clean = normalize_space(line);
// 2. 转小写（保留所有的大写I）
clean = to_lower_custom(clean);
// 3. 替换独立的 I 和 me 为 you
replace_word(clean, "I", "you");
replace_word(clean, "me", "you");
// 4. 替换 can you -> I can, could you -> I could
replace_word(clean, "could you", "I could");
replace_word(clean, "can you", "I can");
// 5. 替换问号为感叹号
for (char& c : clean) {
if (c == '?') c = '!';
}
// 输出 AI 回复
cout << "AI: " << clean << endl;
}
return 0;
}
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