Bovine Birthday（思维题）

Bovine Birthday

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2211 Accepted: 1020

Description

Bessie asked her friend what day of the week she was born on. She knew that she was born on 2003 May 25, but didn't know what day it was. Write a program to help. Note that no cow was born earlier than the year 1800.

Facts to know:

* January 1, 1900 was on a Monday.

* Lengths of months:

    Jan 31          May 31      Sep 30    Feb 28 or 29    Jun 30      Oct 31    Mar 31          Jul 31      Nov 30    Apr 30          Aug 31      Dec 31

* Every year evenly divisible by 4 is a leap year (1992 = 4*498 so 1992 will be a leap year, but the year 1990 is not a leap year).

* The rule above does not hold for century years. Century years divisible by 400 are leap years, all other are not. Thus, the century years 1700, 1800, 1900 and 2100 are not leap years, but 2000 is a leap year.

Input

* Line 1: Three space-separated integers that represent respectively the year, month (range 1..12), and day of a date.

Output

* Line 1: A single word that is the day of the week of the specified date (from the lower-case list: monday, tuesday, wednesday, thursday, friday, saturday, sunday).

Sample Input

2003 5 25

Sample Output

sunday

Hint

INPUT DETAILS:
May 25, 2003

OUTPUT DETAILS:

      May 2003Su Mo Tu We Th Fr Sa             1  2  3 4  5  6  7  8  9 1011 12 13 14 15 16 1718 19 20 21 22 23 2425 26 27 28 29 30 31

题意：

给你某一年1月1日是周一，然后让你随便输入一个日期，判断是周几。

思路：

for循环，对每年每月情况判断。

代码：

#include <iostream>#include <cstring>#include <string>using namespace std;string week[7]={"monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"};bool IsLeapYear(int y)  //判断是否是闰年{    return (y%100!=0&&y%4==0)||(y%400 == 0);}int main(){    int year,month,day;    int i,j,k;    cin>>year>>month>>day;    int w=1;    for(i=1800;i<=2100;i++)    {        for(j=1;j<=12;j++)        {            if(j==4||j==6||j==9||j==11)            {                for(k=1;k<=30;k++)                {                    w=(w+1)%7;                    if(year==i&&j==month&&k==day)cout<<week[w]<<endl;                }            }            else if(j==2)            {                if(IsLeapYear(i))                for(k=1;k<=29;k++){w=(w+1)%7;if(year==i&&j==month&&k==day)cout<<week[w]<<endl;}                else                for(k=1;k<=28;k++){w=(w+1)%7;if(year==i&&j==month&&k==day)cout<<week[w]<<endl;}            }            else            {                for(k=1;k<=31;k++)                {                    w=(w+1)%7;                    if(year==i&&j==month&&k==day)cout<<week[w]<<endl;                }            }        }    }    return 0;}

锻炼敲代码能力。