ZOJ - 3939 The Lucky Week(日期循环节+思维)

最新推荐文章于 2019-04-12 12:44:03 发布

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最新推荐文章于 2019-04-12 12:44:03 发布

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781749.html

本文介绍了一个算法挑战，即计算第N个幸运周的日期。幸运周定义为该月的1日、11日或21日是周一的那周。通过分析四百年周期内特定日期的规律，使用C++实现了一种解决方案，该方案能够处理从1753年至9999年的日期范围。

Edward, the headmaster of the Marjar University, is very busy every day and always forgets the date.

There was one day Edward suddenly found that if Monday was the 1st, 11th or 21st day of that month, he could remember the date clearly in that week. Therefore, he called such week "The Lucky Week".

But now Edward only remembers the date of his first Lucky Week because of the age-related memory loss, and he wants to know the date of the N-th Lucky Week. Can you help him?

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. For each test case:

The only line contains four integers Y, M, D and N (1 ≤ N ≤ 109) indicating the date (Y: year, M: month, D: day) of the Monday of the first Lucky Week and the Edward's query N.

The Monday of the first Lucky Week is between 1st Jan, 1753 and 31st Dec, 9999 (inclusive).

Output

For each case, print the date of the Monday of the N-th Lucky Week.

Sample Input

2
2016 4 11 2
2016 1 11 10

Sample Output

2016 7 11
2017 9 11

这个题的关键在于如何去求一个循环周期的时间

//1:四百年一轮回，从闰年和平年的判定可以推出。
//2:由上一条可以用程序判断出每四百年有2058个天为1，11，21的星期一，直接用。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int a[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int b[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
bool run(int x) {
	if((x%4==0&&x%100!=0)||x%400==0) {
		return true;
	} else {
		return false;
	}
}
int main() {
	int T;
	cin>>T;
	int year  ,month ,day,N;
	while(T--) {
		scanf("%d%d%d%d",&year,&month,&day,&N);
		year +=(N/2058)*400;
		N%=2058;
		int k=day;
		int cnt=0;
		int s=1;

		while(s<=N) {
			if(run(year)) {
				while(day<=b[month]) {

					if((cnt)%7==0&&(day==1||day==11||day==21)) {
						if(s==N)
						{
							cout<<year<<" "<<month<<" "<<day<<endl;
						}
						s++;
					}
					cnt++;
					day++;

				}
				day=1;
				month++;
				if(month>12) {
					month=1;
					year++;
				}
			}


			else {
				while(day<=a[month]) {

					if((cnt)%7==0&&(day==1||day==11||day==21)) {
						if(s==N)
						{
							cout<<year<<" "<<month<<" "<<day<<endl;
						}
						s++;
					}
					cnt++;
					day++;
				}
				day=1;
				month++;
				if(month>12) {
					month=1;
					year++;
				}
			}
		}


	}
	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781749.html