能用标准库写的东西绝不手搓

这题水起来是这样过的（C++11）

/*                     _ooOoo_
                      o8888888o
                      88" . "88
                      (| -_- |)
                       O\ = /O
                   ____/`---'\____
                 .   ' \\| |// `.
                  / \\||| : |||// \
                / _||||| -:- |||||- \
                  | | \\\ - /// | |
                | \_| ''\---/'' | |
                 \ .-\__ `-` ___/-. /
              ___`. .' /--.--\ `. . __
           ."" '< `.___\_<|>_/___.' >'"".
          | | : `- \`.;`\ _ /`;.`/ - ` : | |
            \ \ `-. \_ __\ /__ _/ .-` / /
    ======`-.____`-.___\_____/___.-`____.-'======
                       `=---='                      */
#include <bits/stdc++.h>

using namespace std;
using i128 [[maybe_unused]] = __int128;

std::istream &operator>>(std::istream &is, i128 &value) {
    value = 0;
    bool isNegative = false;
    int c = is.get();
    while (c < '0' or c > '9') {
        if (c == '-')
            isNegative = true;
        c = is.get();
    }
    while (c >= '0' and c <= '9') {
        value = (value << 3) + (value << 1) + (c ^ '0');
        c = is.get();
    }
    if (isNegative)
        value = -value;
    return is;
}

std::ostream &operator<<(std::ostream &os, i128 value) {
    if (value < 0) {
        value = ~value + 1;
        os.put('-');
    }
    static char sta[40];
    char top = 0;
    for (; value; value /= 10)
        sta[top++] = static_cast<char>(value % 10);
    for (; top--;)
        os.put(static_cast<char>(sta[top] ^ '0'));
    return os;
}

using i64 [[maybe_unused]] = long long;
template<typename T, class comparer = greater<T>>
using heap [[maybe_unused]] = priority_queue<T, vector<T>, comparer>;
template<typename T1, typename T2>
using hash_map [[maybe_unused]] = unordered_map<T1, T2>;
template<typename T>
using hash_set [[maybe_unused]] = unordered_set<T>;
const char &ln = '\n';

template<typename T, typename V>
T as [[maybe_unused]](const V &val) { return static_cast<T>(val); }

template<typename T, class Begin, class End>
void println [[maybe_unused]](Begin begin, End end, const char *c = " ") {
    copy(begin, end, ostream_iterator<T>(cout, c));
    cout.put('\n');
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    time_t utc;
    cin >> utc;
    // 微秒转秒，忽略不计了
    utc /= 1000;
    char timeString[size("hh:mm:ss")];
// Windows的MSVC为了安全已经弃用gmtime函数，需要使用更安全的实现。但是评测机是Linux的，所以正常写也无妨
#ifdef _WIN32
    unique_ptr<tm> p(new tm);
    gmtime_s(p.get(), &utc);
#endif
    strftime(data(timeString), size(timeString), "%T",
#ifdef _WIN32
             p.get()
#else
            gmtime(&utc)
#endif
    );
    cout << timeString << ln;
    return 0;
}

看上去很长，其实主要是模板和预处理长，要去掉那些，代码就非常精简了。

#include <bits/stdc++.h>
using namespace std;
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    time_t utc = (cin >> utc, utc / 1000);  // 输入并微秒转秒，去除微妙
    char timeString[9];
    strftime(data(timeString), size(timeString), "%T",gmtime(&utc));
    cout << timeString << '\n';
    return 0;
}

【2021年省赛B组】试题F: 时间显示

题解

简单模拟题，语法补零， 由于不用考虑毫秒，所以可以去掉毫秒的精度。由于不用考虑年月日，所以只保留最后一天的时间即可。值的注意的是$1s=1000s$， 时间复杂度为$O(1)$。

提交代码

• scanf printf 版

#include<bits/stdc++.h>

void solve()
{
    long long x; scanf("%lld", &x);
    x /= 1000;
    x %= 24 * 60 * 60;
    printf("%02lld:%02lld:%02lld", x / 3600, (x % 3600) / 60, (x % 3600) % 60);
}

int main() 
{
    int t = 1;
    while (t--) solve();
    return 0;
}

• cin cout 版

#include<bits/stdc++.h>

void solve()
{
	long long x; std::cin >> x;
    x /= 1000;
    x %= 24 * 60 * 60;
	int hh = x / 3600;
	int mm = (x % 3600) / 60;
	int ss =  (x % 3600) % 60;
    std::cout << std::setfill('0') << std::setw(2) << hh << ":" << std::setw(2) << mm << ":" << std::setw(2) << ss << "\n";
}
int main() 
{
    int t = 1;
    while (t--) solve();
    return 0;
}