ZOJ 3876--May Day Holiday

May Day Holiday

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3876

Description

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

3201520162017

Output

569

题意：已知 2015 年的五一节一号是星期五所以有 5 天假期， 2016 年五月一号是星期天，

            所以除了五天假期还有星期六总共就有 6 天假期， 2017 年五月一号是星期一所以

           总共有 9 天假期，（4月29--5月7）即头尾周末碰巧休假。根据给出的年份求出五

           一假期总天数。

分析：用数组存储周一到周日的五一节假期天数。

        以 2015 年五一为星期五为基准，分输入年份小于 2015 和大于 2015来讨论。

        1：年份大于 2015

              从 2015年5月1日 到 2016年5月1日共经过一年时间，二月份天数主要由 2016 年

             是否为闰年来决定，其它年份依次往上加经过的总天数，最后对 7 取余（一周7天），

             当取余值为 0 时即表示为周日。

       2：年份小于 2015

            恰好是上述情况的逆，假如目前年份五一为星期 k，经过 n 天到 2015年五一星期五。

            则有式子       (k + n % 7) % 7 == 5;     -->  k = 5 + 7 - (n % 7);(加上 7 为了避免结果为负数)

代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <map>#include <functional>#include <algorithm>using namespace std;#define inf 0x3f3f3f3fbool isleap(int y){if ((y % 4 == 0 && y % 100) || (y % 400 == 0))return true;return false;}int num[8] = { 6, 9, 6, 5, 5, 5, 5 };int main(){#ifdef OFFLINEfreopen("t.txt", "r", stdin);#endifint t, i, j, k, y, s;scanf("%d", &t);while (t--){scanf("%d", &y);if (y == 2015) {puts("5"); continue;}k = 5;  s = 0;if (y > 2015){for (i = 2016; i <= y; i++){if (isleap(i))  k += 366;else   k += 365;k %= 7;}k %= 7;}else{for (i = 2015; i > y; i--){if (isleap(i))  s += 366;else   s += 365;}k = k + 7 - (s % 7);k %= 7;}printf("%d\n", num[k]);}return 0;}