Daily exercise: a series of dates

Calculate date to day conversion

link : Calculate date to day conversion

Ideas 1： The conventional idea is to use while Cycle from the first day of this month to that day , But there are still very simple ideas , That is, the following idea 2 , There is also mainly about train of thought 2 .

Train of thought two ： Use an array to store the accumulated days of each month , If the first month is 31 God , The second month is storage 31+28=59 God , And so on , It's good to store like this .
We ask that the total number of days from this year to a certain day is the month before this month , And the sum of all the days before this month plus the days of this month ！ But don't forget to judge whether this year is a leap year and whether this day exceeds February （ Because there is no need to add another day before February ）

Code ：

#include<iostream>
using namespace std;
int main()
{
    
    int arr[13] = {
     0,31,59,90,120,151,181,212,243,273,304,334,365};
    int year, month, day;
    cin >> year >> month >> day;
    
    // Give Way n Assign the total number of days in the previous month from this year to this month plus the current number of days day
    int n = arr[month - 1] + day;
    
    // Remember to judge whether it is a leap year 
    if(month > 2 && (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)))
        n++;
    
    cout << n;
    return 0;
}

Date accumulation

link : Date accumulation

Ideas ： Because this problem requires m Test cases , So cycle m All over . And use nowday Record the maximum number of days in the month , For later judgment .
And then use while Sub loop , take addday The days accumulated to day and month On , then addday–.
Each cycle judges whether the current day exceeds the maximum number of days in the current month , And whether the month exceeds 12 month .

Code ：

#include<iostream>
using namespace std;
int main()
{
    
    int arr[13] = {
    0,31,28,31,30,31,30,31,31,30,31,30,31};
    int m = 0;
    cin >> m;
    int year = 0, month = 1, day = 1, addday = 0;
    
    // Because there are multiple sets of test cases , So use m To cycle 
    for(int i=0; i<m; i++)
    {
    
        cin >> year >> month >> day >> addday;
        
        int nowday = arr[month];// use nowday Record the maximum number of days in the month 
        
        while(addday > 0)
        {
    
            day++;
            if((month == 2) && ((year%4==0 && year%100!=0) || (year%400==0)))
            {
    
                nowday=29;
            }
            
            // Judge whether the number of days is greater than the maximum number of days in the month 
            if(day > nowday)
                {
    
                    day = 1;
                    month++;
                    // Determine whether the number of months is greater than 12
                    if(month > 12)
                    {
    
                        month=1;
                        year++;
                    }
                    nowday=arr[month];
                }
            addday--;
        }
        
    // Print the days and months in format  
    if(day<10 && month<10)
        cout<<year<<"-0"<<month<<"-0"<<day<<endl;
    else if(day<10 && month>=10)
        cout<<year<<'-'<<month<<"-0"<<day<<endl;
    else if(day>=10 && month<10)
        cout<<year<<"-0"<<month<<'-'<<day<<endl;
    else
        cout<<year<<'-'<<month<<'-'<<day<<endl;
    }
    return 0;
}

---

Print date

link : Print date

Ideas ： First use sday Total storage days , And then sday Judge whether this year is a leap year , And then use while The cycle will sday Split into day and month, Every time day++,sday–, until sday be equal to 0.
Then remember to judge day Is the maximum number of days of the month reached , If yes, reset , We also need to judge whether the month is greater than 12, If yes, reset it .

Code ：

#include<iostream>
using namespace std;
int main()
{
    
    int arr[13]={
    0,31,28,31,30,31,30,31,31,30,31,30,31};
    int year=0,month=1,day=0,sday=0;
        cin>>year>>sday;
    
    // Judge whether it is a leap year 
    if((year % 4 ==0 && year % 100 != 0) || (year % 400 == 0))
        arr[2]=29;
    
    // Convert the total number of days into months and days 
    while(sday > 0)
    {
    
        day++;
        if(day > arr[month])
        {
    
            day=1;
            month++;
            if(month > 12)
            {
    
                month=1;
                year++;
            }
        }
        sday--;
    }
        
    // Determine the format of printing days and months 
    if(day<10 && month<10)
        cout<<year<<"-0"<<month<<"-0"<<day<<endl;
    else if(day<10 && month>=10)
        cout<<year<<'-'<<month<<"-0"<<day<<endl;
    else if(day>=10 && month<10)
        cout<<year<<"-0"<<month<<'-'<<day<<endl;
    else
        cout<<year<<'-'<<month<<'-'<<day<<endl;
    
    // Because there are multiple sets of test cases , All to reset 
    arr[2]=28;
    month=day=1;
    
    return 0;
}

---

Date difference

link : Date difference

Ideas ： First save the year, month and day of the two numbers with three variables , keep max Some are big dates ,min For small dates .
And then start the cycle , until min The date of is equal to max Until mm / DD / yyyy , use count To count the number of days between them .
Remember to judge whether the sky exceeds Whether the number of days and months of the month exceeds 12 month , Reset if you have any ！

Code ：

#include<iostream>
using namespace std;
int main()
{
    
    int arr[13]={
    0,31,28,31,30,31,30,31,31,30,31,30,31};
    int day1,day2,count=0;
    cin>>day1>>day2;
    
    // keep max For the big number , Then decompose it 
    int max=day1>day2?day1:day2;
    int maxday=max%100;
    int maxmonth=(max/100)%100;
    int maxyear=max/10000;
    
    // keep min For the small number , Then decompose it 
    int min=day1>day2?day2:day1;
    int minday=min%100;
    int minmonth=(min/100)%100;
    int minyear=min/10000;
    
    // Do not exit the loop until the three values are equal 
    while((minyear!=maxyear)||(minmonth!=maxmonth)||(minday<=maxday))
    {
    
        count++;
        minday++;
        // Judge whether the number of days exceeds the maximum number of days in the month 
        if(minday>arr[minmonth])
        {
    
            minday=1;
            minmonth++;
			// Determine whether the number of months exceeds 12
            if(minmonth>12)
            {
    
                minmonth=1;
                minyear++;
            }
        }
    }
    
    cout<<count;
    return 0;
}

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