[Project Euler] 来做欧拉项目练习题吧: 题目019

                            [Project Euler] 来做欧拉项目练习题吧: 题目019

                                              周银辉 

 

题目描述:

You are given the following information, but you may prefer to do some research for yourself.

• 1 Jan 1900 was a Monday.
• Thirty days has September,
  April, June and November.
  All the rest have thirty-one,
  Saving February alone,
  Which has twenty-eight, rain or shine.
  And on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)? 

 

问题分析:

计算从1901年到2000年间，有多少个星期天是在月初。

题目说1900年1月1日是星期一，也就是说1900年1月7日是星期天，那么从这天开始每增加7天都是星期天。

然后问题就转换成了这些星期天中哪些是在月初。

高级语言都内置了日期函数，如果用类似于date.addDays(n)之类的函数，这个问题就变得非常简单了。

不使用内置函数的话，就用下面的get_days和get_date方法吧:

#include <stdio.h>

long get_days(long y, long m, long d)

{

m = (m + 9)%12;                /* mar=0, feb=11 */

y = y - m/10;                  /* if Jan/Feb, year-- */

return y*365 + y/4 - y/100 + y/400 + (m*306 + 5)/10 + (d - 1);

}

void get_date(long days, long* y, long* m, long* d)

{

long yy, ddd, mm, dd, mi;

yy = (10000*days + 14780)/3652425;

ddd = days - (yy*365 + yy/4 - yy/100 + yy/400);

 

if (ddd < 0) 

{

yy--;

ddd = days - (yy*365 + yy/4 - yy/100 + yy/400);

}

mi = (52 + 100*ddd)/3060;

 

*y = yy + (mi + 2)/12;

*m = (mi + 2)%12 + 1;

*d = ddd - (mi*306 + 5)/10 + 1;

}

int main()

{

long count = 0;

long year  = 1900;

long month;

long day;

long days = get_days(1900, 1, 7);

while(1)

{

days += 7; //增加7天，肯定是星期天

//取得该天的日期，判断其是否是1号

get_date(days, &year, &month, &day);

//printf("%ld-%ld-%ld\n", year, month, day);

if (year>2000) 

{

break;

}

 

if(day==1 && year>=1901)

{

//今天是1号，那么至少要等28天以后才可能是1号了，下次循环是要加7天，所以这里加21天

//避免无用的运算

days  += 21; 

count ++;

}

}

printf("total count %ld\n", count);

return 0;

} 

 

此题几乎不需要多少时间:

real    0m0.005s

user    0m0.001s

sys     0m0.003s