How Many Nines 暴力打表

If we represent a date in the format YYYY-MM-DD (for example, 2017-04-09), do you know how many 9s will appear in all the dates between Y1-M1-D1 and Y2-M2-D2 (both inclusive)?

Note that you should take leap years into consideration. A leap year is a year which can be divided by 400 or can be divided by 4 but can't be divided by 100.

Input

The first line of the input is an integer T (1 ≤ T ≤ 105), indicating the number of test cases. Then T test cases follow. For each test case:

The first and only line contains six integers Y1, M1, D1, Y2, M2, D2, their meanings are described above.

It's guaranteed that Y1-M1-D1 is not larger than Y2-M2-D2. Both Y1-M1-D1 and Y2-M2-D2are between 2000-01-01 and 9999-12-31, and both dates are valid.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each test case, you should output one line containing one integer, indicating the answer of this test case.

Sample Input
4
2017 04 09 2017 05 09
2100 02 01 2100 03 01
9996 02 01 9996 03 01
2000 01 01 9999 12 31
Sample Output
4
2
93
1763534
Hint

For the first test case, four 9s appear in all the dates between 2017-04-09 and 2017-05-09. They are: 2017-04-09 (one 9), 2017-04-19 (one 9), 2017-04-29 (one 9), and 2017-05-09 (one 9).

For the second test case, as year 2100 is not a leap year, only two 9s appear in all the dates between 2100-02-01 and 2100-03-01. They are: 2017-02-09 (one 9) and 2017-02-19 (one 9).

For the third test case, at least three 9s appear in each date between 9996-02-01 and 9996-03-01. Also, there are three additional nines, namely 9996-02-09 (one 9), 9996-02-19 (one 9) and 9996-02-29 (one 9). So the answer is 3 × 30 + 3 = 93.



      题目:给你两个日期,求这两个日期之间九的个数;

      思路:设a[i][j][k]为2000年01月01日到 i 年 j 月 k 日 中 9 的个数,先把所有的情况计算出来,最后用两段日期相减即可。

      代码:

      

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;
int  judge(int x)//判断有多少个9
{
	int n;
	int flag = 0;
	while (x > 0)
	{
		n = x % 10;
		if (n == 9)
			flag++;
		x = x / 10;
	}
	return flag;
}
bool isleap(int x)
{
	if ((x % 4 == 0 && x % 100 != 0) || x % 400 == 0)
		return true;
	else
		return false;
}
int days[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int leapdays[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
int a[10000][13][32];
int main()
{
	int x;
		int t;
		scanf("%d", &t);
		int i, j, k;
		int ans;
		int y1, m1, d1, y2, m2, d2;
		a[2000][1][1] = 0;
		int pre = 0;
		for (i = 2000; i <= 9999; i++)
		{
			if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)
			{
				for (j = 1; j <= 12; j++)
				{
					for (k = 1; k <= leapdays[j]; k++)
					{
						 x = i * 10000 + j * 100 + k;//判断这一天有多少个9
						a[i][j][k] = judge(x) + pre;
						pre = a[i][j][k];
					}
				}
			}
			else
			{
				for (j = 1; j <= 12; j++)
				{
					for (k = 1; k <= days[j]; k++)
					{
                        x = i * 10000 + j * 100 + k;
						a[i][j][k] = judge(x) + pre;
						pre = a[i][j][k];
					}

				}
			}
		}

	for (int i = 0; i < t; i++)
	{
		scanf("%d%d%d", &y1, &m1, &d1);
		scanf("%d%d%d", &y2, &m2, &d2);
		int x = y1 * 10000 + m1 * 100 + d1;
		ans = judge(x);
		printf("%d\n", a[y2][m2][d2] - a[y1][m1][d1]+ans);
	}
	return 0;
}





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