ZOJ3939-The Lucky Week

The Lucky Week
Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

Author: GAN, Tiansheng
Source: The 13th Zhejiang Provincial Collegiate Programming Contest

题意：给一日期，如果这个月的1, 11, 21号是星期一没那么那一天是是幸运的，求所给的日期后第n个幸运天的日期

解题思路：直接打表得到400年为一个循环，且400年内的lucky week的个数为2058，然后暴力求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int mon[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int monn[] = { 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

bool check(int x)
{
if ((x % 4 == 0 && x % 100 != 0) || x % 400 == 0) return 1;
else return 0;
}

int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int y, m, d, n;
scanf("%d %d %d %d", &y, &m, &d, &n);
int yy = n / 2058 * 400 + y;
n %= 2058;
int cnt = 0,sum=0,flag=0;
for (int i = yy; ; i++)
{
for (int j = (i == yy) ? m : 1; j < 13; j++)
{
for (int k = (i == yy && j == m) ? d : 1; k < 30; k += 10,sum+=10)
{
if (sum%7==0) cnt++;
if (cnt == n)
{
flag = 1;
printf("%d %d %d\n", i, j, k);
break;
}
}
if (flag) break;
if (check(i)) sum += monn[j] - 20-10;
else sum += mon[j] - 20-10;
}
if (flag) break;
}
}
return 0;
}