机考指南第二章-经典入门-日期问题自解

日期类问题：

1.基本问题：

求两个日期间的天数差，即求以两个特定日期为界的日期区间的长度。

2.套路：

将原区间问题统一到起点确定的区间问题上去（比如与0000年1月1日的日期之差）。这样处理可以对数据进行预处理，在输入之前就预处理出所有日期与原点日期之间的天数差保存起来，当数据真正开始输入时，只需O(1)的时间复杂度将保存的数据读出即可，是一种空间换时间的重要手段。

3.注意的要点——闰年：

每逢闰年2月将会有29天。闰年的判定——当年数不能被100整除时若其能被4整除则为闰年，或者其能被400整除也是闰年。即Year%100!=0&&Year%4==0||Year%400==0。

例2.3 日期差值

AC代码：

#include<cstdio>
using namespace std;

bool LeapYear(int x)
{
    return (x % 100 != 0 && x % 4 == 0) || x % 400 == 0;
}

int dayOfMonth[13][2] = { 0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31 };

struct Date 
{
    int Day;
    int Month;
    int Year;
    void NextDay()//计算下一天的日期
    {
        Day++;
        if (Day > dayOfMonth[Month][LeapYear(Year)])
        {
            Day = 1;
            Month++;
            if (Month > 12)
            {
                Month = 1;
                Year++;
            }
        }
    }
};

int days[5001][13][32];//保存将要处理的天数

int Abs(int x)//求绝对值
{
    return x < 0 ? -x : x;
}

int main()
{
    Date tmp;
    int cnt = 0;
    tmp.Day = 1;
    tmp.Month = 1;
    tmp.Year = 0;//初始化对象为0000年1月1日
    while (tmp.Year != 5001)//预处理
    {
        days[tmp.Year][tmp.Month][tmp.Day] = cnt;//将该日与0000年1月1日的天数差保存起来
        tmp.NextDay();//计算下一天的日期
        cnt++;//计数器累加，每经过一天计数器加1，表示与原点日期间隔又增加一天
    }
    int d1, m1, y1, d2, m2, y2;
    while (scanf("%4d%2d%2d", &y1, &m1, &d1) != EOF)
    {
        scanf("%4d%2d%2d", &y2, &m2, &d2);
        printf("%d
", Abs(days[y2][m2][d2] - days[y1][m1][d1] + 1));
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

int month[13][2] = { 0, 0, 31,31,29,28,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31 };

bool isLeapYear(int x)
{
    return (x % 4 && !(x % 100)) || (x % 400);
}

int main()
{
    int y1, m1, d1, y2, m2, d2;
    int cnt = 1;
    while (scanf("%4d%2d%2d", &y1, &m1, &d1) != EOF && scanf("%4d%2d%2d", &y2, &m2, &d2) != EOF)
    {
        while (y1 != y2 || m1 != m2 || d1 != d2)
        {
            d1++;
            cnt++;
            if (d1 > month[m1][isLeapYear(y1)])
            {
                d1 = 1;
                m1++;
                if (m1 > 12)
                {
                    m1 = 1;
                    y1++;
                }
            }
        }
        printf("%d
", cnt);
    }
    //system("pause");
    return 0;
}

#include<cstdio>
using namespace std;

int month[13][2] = { 0,0,31,31,29,28,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31 };

bool isLeapYear(int x)
{
    return (x % 4 && !(x % 100)) || x % 400;
}

int count(int y1, int m1, int d1)
{
    int y = 0, m = 1, d = 1, c = 0;
    while (y != y1 || m != m1 || d != d1)
    {
        d++;
        c++;
        if (d > month[m][isLeapYear(y)])
        {
            d = 1;
            m++;
            if (m > 12)
            {
                m = 1;
                y++;
            }
        }
    }
    return c;
}

int main()
{
    int y1, d1, m1;
    int y2, d2, m2;
    while (scanf("%04d%02d%02d", &y1, &m1, &d1) != EOF && scanf("%04d%02d%02d", &y2, &m2, &d2)!=EOF)
    {
        int a = count(y1, m1, d1);
        int b = count(y2, m2, d2);
        int ans;
        if (a > b)ans = a - b;
        else ans = b - a;
        printf("%d
", ans + 1);
    }
    return 0;
}

例2.4 Day of Week

AC代码:

#include<cstdio>
#include<cstring>
using namespace std;

bool LeapYear(int x)
{
    return (x % 100 != 0 && x % 4 == 0) || x % 400 == 0;
}

int dayOfMonth[13][2] = { 0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31 };
char WeekName[7][20] = { "Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday" };
char MonthName[13][20] = { "", "January","February","March","April","May","June","July","August","September","October","November","December" };

struct Date
{
    int Day;
    int Month;
    int Year;
    void NextDay()//计算下一天的日期
    {
        Day++;
        if (Day > dayOfMonth[Month][LeapYear(Year)])
        {
            Day = 1;
            Month++;
            if (Month > 12)
            {
                Month = 1;
                Year++;
            }
        }
    }
};

int days[5001][13][32];//保存将要处理的天数

int Abs(int x)//求绝对值
{
    return x < 0 ? -x : x;
}

int main()
{
    Date tmp;
    int cnt = 0;
    tmp.Day = 1;
    tmp.Month = 1;
    tmp.Year = 0;//初始化对象为0000年1月1日
    while (tmp.Year != 5001)//预处理
    {
        days[tmp.Year][tmp.Month][tmp.Day] = cnt;//将该日与0000年1月1日的天数差保存起来
        tmp.NextDay();//计算下一天的日期
        cnt++;//计数器累加，每经过一天计数器加1，表示与原点日期间隔又增加一天
    }
    int d, m, y;
    char s[20];
    while (scanf("%d%s%d", &d, s, &y) != EOF)
    {
        for (m = 1; m <= 12; m++)
        {
            if (strcmp(s, MonthName[m]) == 0)break;
        }
        int index = days[y][m][d] - days[2019][6][10];//比较对象为2019年6月10日,周一
        printf("%s
", WeekName[(index % 7 + 7) % 7]);//处理复数情况
    }
    return 0;
}

#include<cstdio>
#include<cstring>
using namespace std;

int month[13][2] = { 0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31 };
char Month[13][20] = { " ","January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December" };
char Days[7][20] = { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday","Sunday" };
bool isLeapYear(int x)
{
    return (x % 4 == 0 && x % 100 != 0) || x % 400 == 0;
}

int count(int y1, int m1, int d1)
{
    int y = 1, m = 1, d = 1, c = 0;
    while (y != y1 || m != m1 || d != d1)
    {
        d++;
        c++;
        if (d > month[m][isLeapYear(y)])
        {
            d = 1;
            m++;
            if (m > 12)
            {
                m = 1;
                y++;
            }
        }
    }
    return c;
}

int main()
{
    int y1, d1, m1;
    char m[20];
    while (scanf("%d%s%d", &d1, m, &y1) != EOF)
    {
        for (int i = 1; i < 13; i++)
        {
            if (strcmp(m, Month[i]) == 0)
            {
                m1 = i;
                break;
            }
        }
        int cnt = count(y1, m1, d1);
        int ans = cnt % 7;
        printf("%s
", Days[ans]);
    }
    return 0;
}

注意闰年判断不能使用如下语句：

return (x % 4 && !x % 100) || x % 400;

因为“或”是用于肯定思路的，如图中x=4是就会导致错误判断，因为0||1=1。