D - The Lucky Week

最新推荐文章于 2024-07-15 17:53:06 发布

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本文介绍了一个算法挑战，即计算第N个“幸运周”的确切日期。幸运周定义为某个月份中第一天为星期一且该日为1号、11号或21号的那一周。文章提供了输入输出示例及实现代码。

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).

<h4< dd="">Output

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

<h4< dd="">Sample Input

2
2016 4 11 2
2016 1 11 10

<h4< dd="">Sample Output

2016 7 11
2017 9 11

#include <bits/stdc++.h>
using namespace std;

long long y, m, d, n, m1, yy, d1,x;
int main()
{
    //freopen("c.txt", "w", stdout);
    long long i, j, k;
    long long t;
    scanf("%lld", &t);
    while(t--)
    {
        scanf("%lld %lld %lld %lld", &y, &m, &d, &n);
        m1 = m;yy = y;d1 = d;
        long long sum = 0;
        long long aa=0;
        n--;
        x=n/2058;
        y+=400*x;
        n%=2058;
        while(n)
        {
            long long flag = 0;
            if(y % 4 == 0 && y % 100 != 0 || y % 400 == 0)flag = 1;
            if(m == 5 || m == 7 || m == 10 || m == 12 || m == 8 || m == 1 || m == 3)
            {
                if(d == 1 || d == 11){d += 10; sum += 10;}
                else if(d == 21){d = 1; m++; if(m == 13){y++;m = 1;}sum += 11;}
            }
            else if(m == 4 || m == 6 || m == 9 || m == 11)
            {
                if(d == 1 || d == 11){d += 10; sum+= 10;}
                else if(d == 21){d = 1; m++; if(m == 13){y++;m = 1;}sum += 10;}
            }
            else
            {
                if(d == 1 || d == 11){d += 10; sum+= 10;}
                else if(d == 21){
                        d = 1; m++; if(m == 13){y++;m = 1;}//sum += 10;
                        if(flag == 1)sum += 9;
                        else sum += 8;
                    }
            }
            if(sum % 7 == 0)
            {

                //aa++;cout << y << " " << m << " " << d << " " << aa <<endl;
                n--;
            }
            sum%=7;
        }
        cout << y << " " << m << " " << d << endl;
    }
    return 0;
}