H - May Day Holiday

最新推荐文章于 2020-02-23 11:03:34 发布

七九河开

最新推荐文章于 2020-02-23 11:03:34 发布

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本文链接：https://blog.youkuaiyun.com/sdut_jk17_zhangming/article/details/80034840

本文介绍了一个算法，用于计算Marjar大学每年五一劳动节连续假期的天数。该算法考虑了闰年和平年的区别，并根据实际情况计算出最长可达九天的连续假期。

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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 Tindicating 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

Output

5
6

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
int a[10000];
int main()
{
    int t,n,i;
    a[2015] = 5;
    for(i=2016;i<=9999;i++){
        if(i%4==0&&i%100!=0||i%400==0)
            a[i] = a[i-1]+2;
        else
            a[i] = a[i-1]+1;
        if(a[i]>7)
            a[i] -= 7;
    }
    for(i=2014;i>=1928;i--){
        if((i+1)%4==0&&(i+1)%100!=0||(i+1)%400==0)
            a[i] = a[i+1] - 2;
        else
            a[i] = a[i+1] -1;
        if(a[i]<=0)
            a[i]+=7;
    }
    while(cin>>t){
        while(t--){
            cin>>n;
            if(a[n]==1)
                cout<<"9"<<endl;
            else if(a[n]==7||a[n]==2)
                cout<<"6"<<endl;
            else
                cout<<"5"<<endl;
        }
    }
    return 0;
}