方法1：基姆拉尔森公式
(d + 2m + 3(m + 1) / 5 + y + y / 4 - y / 100 + y / 400 + 1) % 7
- d 为日期，范围是1-31
- m 为月份，范围是3-14，当年的1，2月需处理为上一年的13，14月
- y为年份，当月份为1，2时，y需减一
- 结果为0-6，星期日用0表示，星期一为1，以此类推，星期六为6
方法2：每次加1天，加到目标日期
#include <iostream>
#include <string>
#include <unordered_map>
#include <algorithm>
using namespace std;
string months[13] = {
" ","January","February","March","April","May","June","July","August","September","October","November","December"};
unordered_map<string, int> map_month;
string map_day[8] = {
" ","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
int month_day[2][13] = {
{
0,31,28,31,30,31,30,31,31,30,31,30,31}, //common year
{
0,31,29,31,30,31,30,31,31,30,31,30,31} //leap year
};
int leap(int year);
bool dayJudge(int day1, int month1, int year1, int day2, int month2, int year2);
void printDay(int day, int month, int year);
int main(){
for(int i=1; i<=12; i++) map_month[months[i]] = i;
int day, year;
string month;
while(cin>>day>>month>>year){
printDay(day, map_month[month], year);
}
return 0;
}
int leap(int year){
if((year%4==0 && year%100!=0) || year%400==0) return 1;
else return 0;
}
bool dayJudge(int day1, int month1, int year1, int day2, int month2, int year2){
if(year1!=year2) return year1>year2;
else if(month1!=month2) return month1>month2;
else return day1>day2;
}
void printDay(int day, int month, int year){
//以2001年10月9日是星期2为基准
int base_day=9, base_month=10, base_year=2001;
bool later = true;
if(dayJudge(base_day, base_month, base_year, day, month, year)){
swap(base_day, day);
swap(base_month, month);
swap(base_year, year);
later = false;
}
int day_gap = 0;
while(base_year!=year || base_month!=month || base_day!=day){
base_day++;
if(base_day>month_day[leap(base_year)][base_month]){
base_month++;
base_day = 1;
}
if(base_month>12){
base_year++;
base_month = 1;
}
day_gap++;
}
day_gap %= 7;
int ans = later ? (day_gap<=5 ? 2+day_gap : day_gap-5) : (day_gap<=1 ? 2-day_gap : 9-day_gap);
cout<<map_day[ans]<<endl;
return;
}