思路：此类有关月份的题目可以把月份存在一个数组（列表）里，再用判断是否为闰年就可解决该月天数的问题。然后再标记下几号和周几即可得出答案

C：

#include<iostream.h>

void main()
{
    
      
	 int sum=0, week = 6;//sum:公里数  week:当天为周几,起初2000.1.1为周六
     //从2000到2020
	 for (int year = 2000; year <= 2020; year++)
		{
    
      
			int d[13] = {
    
      0,31,28,31,30,31,30,31,31,30,31,30,31};//数组为每个月份的天数
			if(year%4==0)//2000->2020中的闰年,使闰年2月份天数为29天
				d[2] = 29;
            for (int month = 1; month <= 12; month++)//循环月份
                for (int day = 1; day <= d[month]; day++,week++) //循环天数,再使week++
				{
    
      
                     if(week==8)//若week为8,则重回周一
						 week=1;                     
					 if(day==1 || week==1)//当天为周一或一号时,跑两公里
                         sum+=2;
					 else				//当天不为周一或一号时,跑一公里
                         sum+=1;
                     if(year==2020 && month==10 && day==1)//2020.10.1输出公里数
						cout<<sum<<endl;
                 }
         }
}

Java：

public class GroupC_D11 {
    
      
     public static void main(String[]args){
    
      
         int sum = 0;   //公里数
         int week = 6;  //当天为周几,起初2000.1.1为周六
         for (int year = 2000; year <= 2020; year++) {
    
        //从2000到2020
             int []d = {
    
      0,31,28,31,30,31,30,31,31,30,31,30,31}; //数组为每个月份的天数
             if(year%4==0){
    
       //2000->2020中的闰年
                 d[2] = 29; //使闰年2月份天数为29天
             }
             for (int month = 1; month <= 12; month++) {
    
      //循环月份
                 for (int day = 1; day <= d[month]; day++,week++) {
    
       //循环天数,再使week++
                     if (week==8){
    
        //若week为8,则重回周一
                         week=1;
                     }
                     if(day==1 | week==1){
    
        //当天为周一或一号时,跑两公里
                         sum+=2;
                     }else{
    
                       //当天不为周一或一号时,跑一公里
                         sum+=1;
                     }
                     if(year==2020 & month==10 & day==1){
    
      //2020.10.1输出公里数
                         System.out.println(sum);
                     }
                 }
             }
         }
     }
}

python：

week = 6    #公里数
sum = 0     #当天为周几,起初2000.1.1为周六
for year in range(2000,2021):   #从2000到2020
    d = [0,31,28,31,30,31,30,31,31,30,31,30,31]     #列表为每个月份的天数
    if year%4 == 0: #2000->2020中的闰年,使闰年2月份天数为29天
        d[2] = 29
    for month in range(1,13):   #循环月份
        for day in range(1,d[month]+1): #循环天数
            if week == 8:   #若week为8,则重回周一
                week = 1
            if day == 1 or week == 1:   #当天为周一或一号时,跑两公里
                sum = sum + 2
            else:       #当天不为周一或一号时,跑一公里
                sum = sum + 1
            week = week + 1
            if year == 2020 and month == 10 and day == 1:   #2020.10.1输出公里数
                print(sum)

答案：8879