日期问题

函数模板

int months[13] = {
    0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};

int is_leap(int year)
{
    if (year % 4 == 0 && year % 100 || year % 400 == 0)
        return 1;
    return 0;
}

int get_days(int year, int month)
{
    if (month == 2)
        return months[month] + is_leap(year);
    return months[month];
}

bool check(int date)
{
    int year = date / 1000;
    int month = date % 1000 / 100;
    int days = date % 100;
    if (month < 1 || month > 12) return false;
    if (day < 1 || day > get_days(year, month)) return false;
    return true;
}

输入

int y, m, d;
scanf("%04d%02d%02d", &y, &m, &d);

利用string合并数字

for (int i = 0; i < 1e8; i ++ )
{
    string a = to_string(n);
    string b = to_string(i);
    int d = stoi(a + b);
    if (d % 495 == 0)
    {
        printf("%d\n", i)
        break;
    }
}

// 回文处理
for (int i = 1000; i < 9999; i ++ )
{
    string a = to_string(i);
    string b(a.rbegin(), a.rend());
    int date = stoi(a + b);
    if (date >= date1 && date <= date2 && check(date))
        res ++ ;
}

回文判断

回文数

bool is_Palindrome(int num)
{
    int t = num, ans = 0;
    while (t) {
        ans = ans * 10 + t % 10;
        t /= 10;
    }
    return num == ans;
}

回文串

bool is_Palindrome(char s[])
{
    for (int i = 0, j = 7; i < j; i ++, j -- )
        if (s[i] != s[j])
            return false;
        return true;
}

bool is_Palindrome(string s)
{
    for(int i = 0; i < s.length() / 2; i ++)
    {
        if(s[i] != s[s.length() - 1 - i]) return false;
    }
    return true;
}