ZOJ 3876 May Day Holiday[通俗易懂]

全栈程序员-用户IM • 2022年3月5日下午10:00 • 未分类

大家好，又见面了，我是全栈君。

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

Output

5
6
9

之所以记下这个题，是想提醒一下自己写关于闰年的题目时一定要用闰年规则进行判断，别直接除4！！！

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cmath>
#define ls (u<<1)
#define rs (u<<1|1)
#define maxn 1010
#define ll long long
#define INF 1e9
using namespace std;
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
int d[maxn],c[maxn];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,num;
        scanf("%d",&n);
        int a = 0;
        for(int i=1929;i<=n;i++){
            if((i%4==0 && i%100!=0) || (i%400==0)){
   
   //记得要用闰年规则判断
                a ++;
            }
        }
        int b = n - 1928 - a;
        num = a*366 + b*365;//把从1928年5.1到n年5.1的天数算出来就行
        num %= 7;
        num += 2;
        num %= 7;
        if(num == 0 || num == 2){
            printf("6\n");
        }
        else if(num == 1){
            printf("9\n");
        }
        else if(num>=3&&num<=6){
            printf("5\n");
        }
    }
    return 0;
}