What day is it 用基姆拉尔森计算公式算

What day is it

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7699    Accepted Submission(s): 2189

 

Problem Description

Today is Saturday, 17th Nov,2007. Now, if i tell you a date, can you tell me what day it is ?

 

Input

There are multiply cases.
One line is one case.
There are three integers, year(0<year<10000), month(0<=month<13), day(0<=day<32).

 

Output

Output one line.
if the date is illegal, you should output "illegal". Or, you should output what day it is.

 

Sample Input

2007 11 17

 

Sample Output

Saturday

 

Author

LGX

 

Source

HDU 2007-11 Programming Contest_WarmUp

 

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注意不合法的情况，m，d，y必须都大于1，

基姆拉尔森计算公式：

基姆拉尔森计算公式

W= (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1) mod 7

把一月和二月看成是上一年的十三月和十四月，例：如果是1990-1-1则换算成：1989-13-1来代入公式计算。

我还把周六的英文多加了个s

#include<iostream>
using namespace std;
bool leap(int y)
{
    if(y%4==0&&y%100!=0||y%400==0)
    return true;
    else
    return false;
}
int main()
{
    char a[7][15]={"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
    int y,m,d,i,j,b[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    while(~scanf("%d%d%d",&y,&m,&d))
    {
        if(leap(y))
        b[2]=29;
        if(y>10000||y<1||m<1||m>12||d>b[m]||d<1)
        {
            printf("illegal\n");
            continue;
        }
        if(m<3)
        {
            y--;
            m+=12;
        }
        int w=(d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;    核心在这，基姆拉尔森公式（d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7
        printf("%s\n",a[w]);
        b[2]=28;
    }
    return 0;
}