ZOJ 3939 The Lucky Week 2016省赛

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

唉，这道题模拟的时候没做出来，结束了以后又花了好多时间才做出来了。简直了，果然水题做的还不够多啊。

就是先打表找规律吧，然后就找到1753.1.1是周一，那后其他就还好了

#include

using namespace std;

typedef long long ll;
struct point{
int year, month, day;
int cnt;
point (int year = 0, int month = 0, int day = 0, int cnt = 0): year(year), month(month), day(day), cnt(cnt){}
};
point a[10000];
int n;
int cnt;
int d1[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int d2[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

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

void init(){
cnt = 1;
int day = 1;
for (int i = 1753; i <= 1753 + 399; i++){
bool flag = false;
flag = check(i);
for (int j = 1; j <= 12; j++){
int tmp = (flag == 1 ? d2[j] : d1[j]);
while (day <= tmp){
if (day == 1 || day == 11 || day == 21){
a[cnt] = point(i, j, day, cnt);
cnt++;
}
day += 7;
}
day -= tmp;
}
}
}

int main(){
init();
cnt--;
int t;
scanf("%d", &t);
while (t--){
int y, m, d;
scanf("%d%d%d", &y, &m ,&d);
int n;
scanf("%d", &n);
n--;
int ind = n / cnt;//有几个四百年
int tmp = n - ind * cnt;//除了四百年以后，然后后面还有几个
int mid = 0;//表示目前输入所在的位置在2153年以前的位置，然后看看要加上多少
while (y >= 1753 + 400){
y -= 400;
mid++;
}
//printf("y = %d mid = %d tmp = %d\n", y, mid, tmp);
ll t1 = 0, t2 = 0, t3 = 0;
for (int i = 1; i <= cnt; i++){
if (y == a[i].year && m == a[i].month && d == a[i].day){
int g = i;//就是当前输入-400年以后所在的位置
g += tmp;//表示现在处在的位置
while (g > cnt){
g -= cnt;
mid++;
}
t1 = (ll)a[g].year + (ll)(mid + ind) * 400;
t2 = (ll)a[g].month;
t3 = (ll)a[g].day;
break;
}
}
printf("%lld %lld %lld\n", t1, t2, t3);
}
return 0;
}