洛谷P1202 [USACO1.1]黑色星期五Friday the Thirteenth

题目传送门：                https://www.luogu.com.cn/problem/P1202https://www.luogu.com.cn/problem/P1202

思路：

1、一天一天累加天数

2、如果“星期几”大于7，回到星期一

3、打表每个月天数，有平年和闰年两个表

4、超出这个月天数后月份加一，日期归一，就是说该下一个月了

      这时，要判断闰年，利用打的表确定该月天数

5、超出十二个月后，年份加一，月份归一，要继续新的一年了

6、每到13日，周几的数组就加一

7、到指定年份后停止，再指定年份的下一年一月一日结束

8、最后，按日、一、二、三、四、五、六的顺序输出即可

上代码：

#include <bits/stdc++.h>

using namespace std;

int n, day, month, year, week;
bool ch;
int weekday[8];
int form[2][13] = {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};

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

int main () {
	
	cin >> n;
	
	year = 1900; day = month = week = 1;
	
	while (!(year == 1900 + n && month == 1 && day == 1)) {
		if (f(year)) ch = true;
		else ch = false;
		day++; week++; 
		if (day > form[((ch)?1:0)][month]) {
			day = 1; month++;
		}
		if (month > 12) {
			month = 1; year++;
		}
		if (week > 7) week = 1;
		if (day == 13) weekday[week]++;
	}
	
	cout << weekday[6] << " " << weekday[7] << " " << weekday[1] << " " << weekday[2] << " " << weekday[3] << " " << weekday[4] << " " << weekday[5];
	
	return 0;
}