帮助改善问题

我正在尝试解决这个问题：

有几种完全不同的尝试来解决这个问题：

我首先使用 itertools.groupby 在 python 中编写

from itertools import groupby
from sys import stdin
keys = input().split()
repeat_counts = '-c' in keys
only_repeats = '-d' in keys
ignore_reg = '-i' in keys
only_uniq = '-u' in keys
args = {}
if ignore_reg:
    args = {'key': str.lower}

for st in groupby(map(str.strip, stdin.readlines()), **args):
    counts = sum(1 for _ in st[1])
    if only_repeats and only_uniq:
        continue
    if repeat_counts:
        if only_repeats:
            if counts > 1:
                print(counts, st[0])
        elif only_uniq:
            if counts == 1:
                print(1, st[0])
        else:
            print(counts, st[0])
    else:
        if only_repeats:
            if counts > 1:
                print(st[0])
        elif only_uniq:
            if counts == 1:
                print(st[0])
        else:
            print(st[0])

该任务通过了 11 次测试。

然后我决定用 C++ 编写任何我能用笨拙的方法吃的代码。也就是计算重复次数。

#include <iostream>
#include <sstream>
#include <algorithm>

using namespace std;
bool
    repeat_count = false,
    only_repeats = false,
    without_reg = false,
    only_unique = false;

char change_case(char c) {
    if (isupper(c))
        return tolower(c);
    return c;
}

string pass(string st) {
    return st;
}

string lower(string st) {
    transform(st.begin(), st.end(), st.begin(), change_case);
    return st;
}
void answer(int count, string old){
    if (only_repeats) {
        if (count > 1) {
            if (repeat_count) {
                cout << count << " " << old << '\n';
            } else
            {
                cout << old << "\n";
            }
        }

    } else if (only_unique) {
        if (count == 1) {
            if (repeat_count) {
                cout << 1 << " " << old << "\n";
            }
            else {
                cout << old << "\n";
            }
        }
    }
    else {
        if (repeat_count) {
            cout << count << " " << old << "\n";
        }
        else {
            cout << old << "\n";
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    string keys, key, st;
    getline(cin, keys);
    istringstream ss(keys);
    while (ss >> key) {
        st = key.c_str();
        if (st == "-c") {
            repeat_count = true;
        } else if (st == "-d") {
            only_repeats = true;
        } else if (st == "-i") {
            without_reg = true;
        } else if (st == "-u") {
            only_unique = true;
        }
    }
    if (only_unique and only_repeats)
        return 0;
    string old, cur;
    cin >> old;
    int count = 1;
    auto func = pass;
    if (without_reg) {
        func = lower;
    }
    while (cin >> cur) {
        if (func(cur) == func(old)) {
            count++;
        }
        else {
            answer(count, old);
            count = 1;
            old = cur;
        }
    }
    answer(count, old);

}

任务到了10分。然后我想在12月解决它，在我看来这是一个更可靠的方法。她看起来像这样：

#include "iostream"
#include "algorithm"
#include "sstream"
#include "deque"
using namespace std;

char change_case(char c) {
    return isupper(c) ? tolower(c) : c;
}

string pass(string st) {
    return st;
}

string lower(string st) {
    transform(st.begin(), st.end(), st.begin(), change_case);
    return st;
}

bool
        repeat_count = false,
        only_repeats = false,
        without_reg = false,
        only_unique = false;

auto func = pass;

void init() {
    string keys, key, temp;
    getline(cin, keys);
    istringstream ss(keys);
    while (ss >> key) {
        temp = key.c_str();
        if (temp == "-c") {
            repeat_count = true;
        }
        else if (temp == "-d") {
            only_repeats = true;
        }
        else if (temp == "-i") {
            without_reg = true;
        }
        else if (temp == "-u") {
            only_unique = true;
        }
    }
    if (without_reg)
        func = lower;
}

void answer(const deque <string> &a){
    string out;
    int sz = a.size();
    if (repeat_count){
        if (only_unique){
            if (sz == 1){
                cout << sz << " " << a.front() << "\n";
            }
        }
        else if (only_repeats){
            if (sz != 1){
                cout << sz << " " << a.front() << "\n";
            }
        }
        else{
            cout << sz << " " << a.front() << "\n";
        }
    }
    else{
        if (only_unique){
            if (sz == 1){
                cout << a.front() << "\n";
            }
        }
        else if (only_repeats){
            if (sz != 1){
                cout << a.front() << "\n";
            }
        }
        else{
            cout << a.front() << "\n";
        }
    }
}

int main() {
    init();
    if (only_unique and only_repeats)
        return 0;
    string cur;
    cin >> cur;
    deque <string> st = {cur};
    while (cin >> cur) {
        if (func(cur) == func(st.front())){
            st.push_back(cur);
        }
        else{
            answer(st);
            st.clear();
            st.push_back(cur);
        }
    }
    answer(st);
}

然后她在第 10 次测试中摔倒了。显然，在这两种解决方案中，我忘记考虑的一些小事都存在错误。你能告诉我需要修复/更换什么吗？