九度OJ 1043：Day of Week（星期几） （日期计算）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：5349

解决：1923

题目描述：

We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.
For example, years 2004, 2180 and 2400 are leap. Years 2004, 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.

输入：

There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.

输出：

Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.

样例输入：

9 October 2001
14 October 2001

样例输出：

Tuesday
Sunday

提示：

Month and Week name in Input/Output:
January, February, March, April, May, June, July, August, September, October, November, December
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

来源：

2008年上海交通大学计算机研究生机试真题

思路：

日期计算类的题目不少，虽然不难，但容易出错。

需要注意的地方主要是闰年的计算。

一般的年是365天，二月是28天，而闰年则366天，2月是29天。

闰年的划定标准是：400的倍数，或者4的倍数但不是100的倍数。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define N 10

int compare(int y[2], int m[2], int d[2])
{
    if (y[0] != y[1])
        return y[0]-y[1];
    else if (m[0] != m[1])
        return m[0]-m[1];
    else if (d[0] != d[1])
        return d[0]-d[1];
    else
        return 0;
}

void swap(int a[2])
{
    int tmp;
    tmp = a[0];
    a[0] = a[1];
    a[1] = tmp;
}

int days(int y, int m, int d)
{
    int count = 0;

    //printf("y=%d, m=%d, d=%d\n", y, m, d);

    count += y*365;
    count += (y-1)/4+1;
    count -= (y-1)/100+1;
    count += (y-1)/400+1;
    //printf("count=%d\n", count);

    if (m > 1)
        count += 31;
    if (m > 2)
    {
        if ((y%4 == 0 && y%100 != 0) || y%400 == 0)
            count += 29;
        else
            count += 28;
    }
    if (m > 3)
        count += 31;
    if (m > 4)
        count += 30;
    if (m > 5)
        count += 31;
    if (m > 6)
        count += 30;
    if (m > 7)
        count += 31;
    if (m > 8)
        count += 31;
    if (m > 9)
        count += 30;
    if (m > 10)
        count += 31;
    if (m > 11)
        count += 30;
    if (m > 12)
        count += 31;
    //printf("count=%d\n", count);

    count += d;
    //printf("count=%d\n", count);

    return count;
}

int month(char s[])
{
    char a[12][20] = {"January", "February", "March", "April",
                    "May", "June", "July", "August",
                    "September", "October", "November", "December"};
    int i;
    for (i=0; i<12; i++)
    {
        if (strcmp(s, a[i]) == 0)
            break;
    }   
    return i+1;
}   

void pweek(int w1)
{   
    char s[7][10] = {"Sunday", "Monday", "Tuesday", "Wednesday",
                    "Thursday", "Friday", "Saturday"};
    printf("%s\n", s[w1]);
}   

int main(void)
{       
    char s[N];
    int y[2], m[2], d[2], w[2];

    y[0] = 2001; 
    m[0] = 10;
    d[0] = 9; 
    w[0] = 2;
    while (scanf("%d%s%d", &d[1], s, &y[1]) != EOF)
    {
        m[1] = month(s);
        w[1] = w[0];
        int cmp = compare(y, m, d);
        if (cmp < 0)
        {
            w[1] = (w[0] + days(y[1], m[1], d[1])
                    - days(y[0], m[0], d[0])) % 7;
        }
        else if (cmp > 0)
        {
            w[1] = (w[0] + 7 - (days(y[0], m[0], d[0])
                    - days(y[1], m[1], d[1])) % 7) % 7;
        }           
        pweek(w[1]);
    }

    return 0;
}       
/**************************************************************
    Problem: 1043
    User: liangrx06
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:916 kb
****************************************************************/