Codeforces Gym101522 D.Distribution of Days-算日期 (La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017)
The Gregorian calendar is internationally the most widely used civil calendar. It is named after Pope Gregory XIII, who introduced it in October 1582.
In the Gregorian calendar, there are 28 days in February in a common year and 29days in February in a leap year. Year Y is a leap year if and only if Y is a multiple of 400, or Y is a multiple of 4 and is not a multiple of 100.
Percy is curious about the distribution of days of the week of his birthday in his life. By checking the calendar, he quickly finds that in the years between 1999 and 2017 (inclusive), his birthday (in case you do not know, 27 February) appears only twice on both Tuesday and Thursday, three times on each of the other days of the week.
Percy finds counting the distribution of some days in some consecutive years really cool, so he decides to invent a way to quickly count the distribution.
Within 15 minutes, he successfully invented a fast program to do the calculation for years between 1583 and 2 × 109, inclusive. His program can answer 5000 queries in 1second. However, he is not sure if the program works correctly, so he needs your help. Your task is simple, write your own program to do the calculation, so that Percy can check his program's correctness by comparing the outputs of different queries with your program.
In this problem, please assume the definition of leap years mentioned above is true for all years between 1583 and 2 × 109, inclusive.
Input
The first line consists of a single integer, Q, denotes the number of queries. (1 ≤ Q ≤ 5000)
In the next Q lines, each describes a single query. The queries are in the format S E M D, which means you have to calculate the distribution of days of the week for the D-th day of the M-th month for all years between S and E, inclusive. (1583 ≤ S ≤ E ≤ 2 × 109, the days given are one of the 366 valid days)
Output
Output Q lines, each answers a query given.
In each line output 7 integers, the frequencies of days of the weeks in this order: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
The order of answers should follow the order of queries given.
Example
1
1999 2017 2 27
3 3 2 3 2 3 3
2
2017 2017 8 15
2017 2021 2 29
0 0 1 0 0 0 0
0 0 0 0 0 0 1
4
3141 5926 5 3
5897 9323 8 4
2718 2818 2 8
2222 2222 2 22
404 391 403 390 404 396 398
488 488 497 481 497 480 496
15 14 14 15 14 15 14
0 0 0 0 0 1 0
这个题根本就没有什么技术含量,但是我写T了,
基姆拉尔森计算公式不好用,慎用。
正解(不是我写的,我的T了,不想改了)
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<cmath>
6 #include<stack>
7 #include<map>
8 #include<vector>
9 #include<set>
10 #include<queue>
11 using namespace std;
12 typedef long long ll;
13 const int MAXN=1e5+10;
14 const double mod=1e16+7;
15 int isr(int n)
16 {
17 if(n%4==0&&n%100!=0||n%400==0)return 1;
18 else return 0;
19 }
20 int da[810][15][35];
21
22 void init()
23 {
24 int k[15],l=6;
25 k[1]=31,k[2]=28,k[3]=31,k[4]=30,k[5]=31,k[6]=30;
26 k[7]=31,k[8]=31,k[9]=30,k[10]=31,k[11]=30,k[12]=31;
27 memset(da,0,sizeof(da));
28 for(int i=0;i<=399;i++){
29 for(int j=1;j<=12;j++){
30 if(j==2)
31 for(int t=1;t<=k[j]+isr(i);t++){
32 if(l+1==8)l=0;
33 da[i][j][t]=++l,da[i+400][j][t]=l;
34 }
35 else
36 for(int t=1;t<=k[j];t++){
37 if(l+1==8)l=0;
38 da[i][j][t]=++l,da[i+400][j][t]=l;
39 }
40 }
41 }
42 }
43
44 int main()
45 {
46 int q,y1,y2,m,d,t;
47 ll s[10];
48 cin>>q;
49 init();
50 while(q--){
51 memset(s,0,sizeof(s));
52 scanf("%d%d%d%d",&y1,&y2,&m,&d);
53 if(y2-y1>=400){
54 t=(y2-y1)/400;
55 for(int i=1;i<=400;i++)
56 s[da[i][m][d]]+=t;
57 }
58 y1%=400,y2%=400;
59 if(y1>y2) y2+=400;
60 for(int i=y1;i<=y2;i++){
61 s[da[i][m][d]]++;
62 }
63 for(int i=1;i<=7;i++)
64 cout<<s[i]<<" ";
65 cout<<endl;
66 }
67 }
贴一下我的超时垃圾代码:
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<cmath>
5 #include<math.h>
6 using namespace std;
7 int CaculateWeekDay(int y,int m, int d){
8 if(m==1||m==2){
9 m+=12;
10 y--;
11 }
12 int iWeek=(d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;//基姆拉尔森计算公式根据日期判断星期几
13 if(m==2&&d==29) return iWeek+15;
14 return iWeek+1;
15 }
16 int b[30];
17 int year1=0,year2=0,month=0,day=0;
18 void isbb(){
19 memset(b,0,sizeof(b));
20 for(int i=1;i<=400;i++){
21 int temp=CaculateWeekDay(i,month,day);
22 b[temp]++;
23 }
24 }
25 int main(){
26 int a[10];
27 int t;
28 while(~scanf("%d",&t)){
29 while(t--){
30 scanf("%d%d%d%d",&year1,&year2,&month,&day);
31 memset(a,0,sizeof(a));
32 isbb();
33 if((year2-year1)/400){
34 int x=(year2-year1)/400;
35 if(month==2&&day==29){
36 int j=1;
37 for(int i=15;i<=22;i++)
38 a[j++]=b[i];
39 }
40 else{
41 for(int i=1;i<=7;i++)
42 a[i]=b[i];
43 }
44 for(int i=1;i<=7;i++)
45 a[i]*=x;
46 year1%=400;
47 if(year2%400<year1)year2=year2%400+400;
48 else year2%=400;
49 for(int i=year1;i<=year2;i++){
50 if(month==2&&day==29){
51 if(i%400==0){
52 int temp=CaculateWeekDay(i,month,day);
53 a[temp-15]++;
54 }
55 }
56 else{
57 int temp=CaculateWeekDay(i,month,day);
58 a[temp]++;
59 }
60 }
61 }
62 else{
63 for(int i=year1;i<=year2;i++){
64 int temp=CaculateWeekDay(i,month,day);
65 a[temp]++;
66 }
67 }
68 }
69 cout<<a[7];
70 for(int i=1;i<=6;i++)
71 cout<<" "<<a[i];
72 cout<<endl;
73 }
74 return 0;
75 }
心累。。。
Codeforces Gym101522 D.Distribution of Days-算日期 (La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017)的更多相关文章
- Codeforces Gym101522 C.Cheering-字符串 (La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017)
C.Cheering To boost contestants' performances in the 20th La Salle - Pui Ching Programming Challenge ...
- Codeforces Gym101522 A. Ambiguous Dates (La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017)
A. Ambiguous Dates There are two popular formats for representing a date: day/month/year or month/da ...
- gym101522 [小熊骑士限定]La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017
西瓜队(划掉),Kuma Rider久违的第一场训练,四小时瞎打.jpg A.水题,排序 #include<cstdio> #include<iostream> #includ ...
- Codeforces 1167C - News Distribution
题目链接:http://codeforces.com/problemset/problem/1167/C 题意:大概就是分成几个小团体,给每个人用1 - n编号,当对某个人传播消息的时候,整个小团体就 ...
- Codeforces 1090A - Company Merging - [签到水题][2018-2019 Russia Open High School Programming Contest Problem A]
题目链接:https://codeforces.com/contest/1090/problem/A A conglomerate consists of n companies. To make m ...
- 【Teradata】日期类型计算
1.EXTRACT(抽取年/月/日/时/分/秒) //抽取年/月/日/时/分/秒 SELECT EXTRACT (YEAR FROM CURRENT_DATE); SELECT EXTRACT (M ...
- Python datetime time 等时间 日期 之间的计算和相互转化
from datetime import datetime, date, timedelta, timezone from time import time, ctime, localtime, st ...
- Java8新特性--日期和时间API
如何正确处理时间 现实生活的世界里,时间是不断向前的,如果向前追溯时间的起点,可能是宇宙出生时,又或是是宇宙出现之前, 但肯定是我们目前无法找到的,我们不知道现在距离时间原点的精确距离.所以我们要表示 ...
- CodeForces 164 B. Ancient Berland Hieroglyphs 单调队列
B. Ancient Berland Hieroglyphs 题目连接: http://codeforces.com/problemset/problem/164/B Descriptionww.co ...
随机推荐
- 动态规划、记忆化搜索:HDU1978-How many ways
Problem Description 这是一个简单的生存游戏,你控制一个机器人从一个棋盘的起始点(1,1)走到棋盘的终点(n,m).游戏的规则描述如下: 1.机器人一开始在棋盘的起始点并有起始点所标 ...
- Linux 用户行为日志记录
工作中我们常常遇到,有的员工不安于被分配的权限,老是想sudo echo "ziji" /usr/bin/visudo NOPASSWD:ALL来进行提权,造成误删了数据库某条重要 ...
- CodeForces 781D Axel and Marston in Bitland DP
题意: 有一个\(n\)个点\(m\)条边的无向图,边有两种类型,分别用\(0\)和\(1\)标识 因此图中的任意一条路径都对应一个\(01\)字符串 定义一个无限长的字符串\(s\): 开始令\(s ...
- Redis实现之压缩列表
压缩列表 压缩列表(ziplist)是列表键和哈希键的底层实现之一,当一个列表键只包含少量列表项,并且每个列表项要嘛是整数值,要嘛是比较短的字符串,那么Redis就会使用压缩列表来做列表键的底层实现. ...
- asp.net实现调用ffmpeg实现视频格式的转换
视频格式转换的函数 //视频转换 public void VideoConvertFlv(string FromName, string ExportName) { string ffmpeg = H ...
- Python框架之Django学习笔记(六)
模板 上篇博文学习了动态视图,但是,视图中返回文本的方式有点特别. 也就是说,HTML被直接硬编码在 Python 代码之中. def current_datetime(request): now = ...
- java setVisible顺序不同导致窗体内容不显示问题
今天学习JAVA编写窗体的时候,先写了setVisible(true);然后才去创建的各种控件以及设置大小.位置等 结果运行后只显示空白的窗体,必须最小化再最大化或点击一下边框,才显示窗体内容(即必须 ...
- Leetcode 525.连续数组
连续数组 给定一个二进制数组, 找到含有相同数量的 0 和 1 的最长连续子数组. 示例 1: 输入: [0,1] 输出: 2 说明: [0, 1] 是具有相同数量0和1的最长连续子数组. 示例 2: ...
- 【转】Unity3d的单例及场景间的数据传递
http://blog.csdn.net/zy19940906/article/details/47724387 单例是场景间切换时传递数据的最常见的方式之一,在unity中,很多方法被封装,有时候 ...
- 【Luogu】P3750分手是祝愿(期望DP)
题目链接 这题好喵啊…… 设f[i]是最少用i次才能全关上转移到最少用i-1次才能全关上灯的期望值,那么n个灯里有i个是正确的,剩下的都是不正确的 因此期望是$f[i]=frac{n}{i}+frac ...