题面：
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-D2 are 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.
嘛，实际上是一道比较简单的模拟题……
题意就是给两个日期 年月日 求两个日期间一共有多少9
思路就是 先预处理一下大数组nine[10000][13][32]
定义数组里的值：从2000-01-01 到 nowyear-nowmonth-nowday的9的数目
求9的数目 就是/10 %10 一位位的看就好
初始化好数组之后直接O(1)查就好
Owo 前缀和！ 就是那种 跟数组区间有关系的问题
用前缀和来讲是比较方便的=-=