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)?

#include <stdio.h>
#include <stdbool.h> const int a[][] = {{,,,,,,,,,,,},
{,,,,,,,,,,,}}; bool leapYear(int n) //判断闰年
{
return (((n % ==) && (n % !=)) || (n % == ));
} bool issunday(int n) //判断某天是否是星期天
{
return (n % == ? true : false);
} void solve(void)
{
int num, i, j, count;
count = ; i = ;
num = ;
while(i < ) { int t = (leapYear(i) ? : ); //判断闰年
for(j = ; j < ; j++) {
num += a[t][j];
if(issunday(num)) count++;
}
i++;
}
printf("%d\n",count);
} int main(void)
{
solve();
return ;
}
Answer:
171

(Problem 19)Counting Sundays的更多相关文章

  1. project euler 19: Counting Sundays

    import datetime count = 0 for y in range(1901,2001): for m in range(1,13): if datetime.datetime(y,m, ...

  2. Project Euler 19 Counting Sundays( 蔡勒公式计算星期数 )

    题意:在二十世纪(1901年1月1日到2000年12月31日)中,有多少个月的1号是星期天? 蔡勒公式:计算 ( year , month , day ) 是星期几 以下图片仅供学习! /****** ...

  3. Problem 19

    Problem 19 You are given the following information, but you may prefer to do some research for yours ...

  4. Project Euler:Problem 76 Counting summations

    It is possible to write five as a sum in exactly six different ways: 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 ...

  5. 【DFS深搜初步】HDOJ-2952 Counting Sheep、NYOJ-27 水池数目

    [题目链接:HDOJ-2952] Counting Sheep Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 ...

  6. HDU-2952 Counting Sheep (DFS)

    Counting Sheep Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Tota ...

  7. hdu Counting Sheepsuanga

    算法:深搜 题意:让你判断一共有几个羊圈: 思路:像四个方向搜索: Problem Description A while ago I had trouble sleeping. I used to ...

  8. UVA - 10574 Counting Rectangles

    Description Problem H Counting Rectangles Input: Standard Input Output:Standard Output Time Limit: 3 ...

  9. HDU 2952 Counting Sheep(DFS)

    题目链接 Problem Description A while ago I had trouble sleeping. I used to lie awake, staring at the cei ...

随机推荐

  1. Oracle视图,序列及同义词、集合操作

    一.视图(重点) 视同的功能:一个视图其实就是封装了一个复杂的查询语句.1.创建视图的语法:CREATE VIEW 视图名称 AS 子查询 范例:创建一个包含了20部门的视图CREATE VIEW e ...

  2. leetcode Valid Sudoku python

    #数独(すうどく,Sūdoku)是一种运用纸.笔进行演算的逻辑游戏.玩家需要根据9×9盘面上的已知数字,推理出所有剩余空格的数字,并满足每一行.每一列.每一个粗线宫内的数字均含1-9,不重复.#数独盘 ...

  3. PHP搭建简单暴力的mvc

    对于一个web系统来说,我们使用mvc很必要, 给我们带来的是清晰的结构,易运维,易扩展, mvc 我对其的理解应该叫mxvc, 多了一个x , 这个x代表什么,x可以理解为 relay,proxy, ...

  4. QF——iOS第三方登录和社会化分享

    QQ登录的流程: 1.下载SDK,并添加到项目中: 2.添加SDK需要的依赖库,以及配置文件: 3.重写APPDelegate的方法handleOpenURL和openURL: 4.实现Tencent ...

  5. Customizing Zend Studio Using the Welcome Page

    Customizing Zend Studio Using the Welcome Page Zend Studio enables you to add or remove plugins from ...

  6. MediaStore

    Class Overview 提供的多媒体数据包括内部和扩展的所有多媒体元数据. Summary Nested Classes MediaStore.Audio:此类包含了所有音频相关信息. Medi ...

  7. js、css、html判断浏览器的各种版本

    利用正则表达式来判断ie浏览器版本 判断是否IE浏览器 if (document.all) { alert("这个是ie浏览器");} 判断是否IE6浏览器 方法一:if ( /M ...

  8. 环境配置与JBoss安装-EJB3.0入门经典学习笔记(1)

    目录 1. JDK的安装 2. JBoss的安装 3. JBoss安装目录说明 1. JDK的安装 1) 下载JDK 下载地址:http://www.oracle.com/technetwork/ja ...

  9. 关于Thinkphp3.2版本的分页问题

    最近公司官网改版,需要把旧的thinkphp版本换到现在最新的3.2去,因此,就开展了一系列的升级工作..在修改命名空间的同时,发现Page分页类能正常运行了,但是分页的链接却是错误的,例如在Admi ...

  10. QT连接mysql中文显示问题

    亲测OK! #vim /etc/mysql/my.cnf [mysqld]下面加入: default-character-set=utf8 重启mysql /etc/init.d/mysql rest ...