12004：十亿秒后的时刻

题目

按照无时区公历规则，计算给定时刻十亿秒后的日期与时间。

解析

十亿秒可以先拆成整天和日内余数：

1000000000 = 11574 × 86400 + 6400

先把 6400 秒加入日内时刻，并把可能产生的一天进位并入待增加天数；再按照每月实际天数推进日期。这样不依赖 time_t 的表示，也不会引入本地时区和夏令时规则。

输出对象只在全部计算成功后更新。若日期非法，或跨年时会超过 int64_t 的表示范围，直接返回失败。

解析

#include <stdbool.h>
#include <stdint.h>

typedef struct {
    int64_t year;
    unsigned month;
    unsigned day;
    unsigned hour;
    unsigned minute;
    unsigned second;
} CivilTime;

static bool is_leap_year(int64_t year) {
    return year % 4 == 0 &&
        (year % 100 != 0 || year % 400 == 0);
}

static unsigned days_in_month(
    int64_t year,
    unsigned month
) {
    static const unsigned days[] = {
        31, 28, 31, 30, 31, 30,
        31, 31, 30, 31, 30, 31
    };

    if (month == 2 && is_leap_year(year)) {
        return 29;
    }
    return days[month - 1];
}

static bool civil_time_valid(CivilTime value) {
    if (value.year < 1 ||
        value.month < 1 ||
        value.month > 12 ||
        value.hour > 23 ||
        value.minute > 59 ||
        value.second > 59) {
        return false;
    }

    return value.day >= 1 &&
        value.day <= days_in_month(
            value.year,
            value.month
        );
}

bool add_gigasecond(CivilTime start, CivilTime *out) {
    if (out == NULL || !civil_time_valid(start)) {
        return false;
    }

    const uint32_t seconds_per_day = UINT32_C(86400);
    const uint32_t whole_days = UINT32_C(11574);
    const uint32_t remaining_seconds = UINT32_C(6400);

    CivilTime result = start;
    uint32_t seconds_of_day =
        (uint32_t)result.hour * UINT32_C(3600) +
        (uint32_t)result.minute * UINT32_C(60) +
        result.second +
        remaining_seconds;

    uint32_t days_to_add =
        whole_days + seconds_of_day / seconds_per_day;
    seconds_of_day %= seconds_per_day;

    result.hour = (unsigned)(seconds_of_day / UINT32_C(3600));
    seconds_of_day %= UINT32_C(3600);
    result.minute = (unsigned)(seconds_of_day / UINT32_C(60));
    result.second = (unsigned)(seconds_of_day % UINT32_C(60));

    while (days_to_add > 0) {
        unsigned month_days = days_in_month(
            result.year,
            result.month
        );
        unsigned remaining = month_days - result.day;

        if (days_to_add <= remaining) {
            result.day += (unsigned)days_to_add;
            days_to_add = 0;
            continue;
        }

        days_to_add -= remaining + 1;
        result.day = 1;

        if (result.month < 12) {
            ++result.month;
        } else {
            if (result.year == INT64_MAX) {
                return false;
            }
            result.month = 1;
            ++result.year;
        }
    }

    *out = result;
    return true;
}

来源与改编

题型改编自 Exercism C Track 的 Gigasecond，采用 MIT 许可；本题改为无时区公历接口，并补充日期校验与年份溢出语义。