As a university advocating self-learning and work-rest balance, Marjar University has so many days of rest, including holidays and weekends. Each weekend, which consists of Saturday and Sunday, is a rest time in the Marjar University.
The May Day, also known as International Workers' Day or International Labour Day, falls on May 1st. In Marjar University, the May Day holiday is a five-day vacation from May 1st to May 5th. Due to Saturday or Sunday may be adjacent to the May Day holiday, the continuous vacation may be as long as nine days in reality. For example, the May Day in 2015 is Friday so the continuous vacation is only 5 days (May 1st to May 5th). And the May Day in 2016 is Sunday so the continuous vacation is 6 days (April 30th to May 5th). In 2017, the May Day is Monday so the vacation is 9 days (April 29th to May 7th). How excited!
Edward, the headmaster of Marjar University, is very curious how long is the continuous vacation containing May Day in different years. Can you help him?
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case, there is an integer y (1928 <= y <= 9999) in one line, indicating the year of Edward's query.
Output
For each case, print the number of days of the continuous vacation in that year.
Sample Input
3 2015 2016 2017
Output
5 6 9
Author: ZHOU, Yuchen
Source: The 12th Zhejiang Provincial Collegiate Programming Contest
问题链接：ZOJ3876 May Day Holiday。
问题简述：参见上文，根据输入的年份，计算五一节连休几天：
问题分析：（略）
程序说明：
需要知道起始年份（1928年）5月1日是星期几？需要编写程序计算。根据题意已知，2015年5月1日是星期五，据此推算。结果是1928年5月1日是星期二。
用0-6分别表示星期天到星期六。
先计算5月1日分别是星期几（0-6）时，连休的天数，并且放在数组ans[]中备查。
根据起始年份，逐年推算5月1日是星期几就可以了。
函数leapyear()用于计算是否为闰年（闰年则为1，否则为0）。
AC的C++语言程序如下：
/* ZOJ3876 May Day Holiday */
#include <iostream>
#include <stdio.h>
using namespace std;
int ans[] = {6, 9, 6, 5, 5, 5, 5};
int leapyear(int year)
{
return ( ((year%4==0) && (year%100!=0)) || (year%400==0) ) ? 1 : 0;
}
int main()
{
int t, year;
scanf("%d", &t);
while(t--) {
scanf("%d", &year);
int d=2, y=1928;
while(year != y) {
y++;
d = (d + 365 + leapyear(y)) % 7;
}
printf("%d
", ans[d]);
}
return 0;
}