Arrange
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1191 Accepted Submission(s): 404
Problem Description
Accidentally, Cupid, god of desire has hurt himself with his own dart and fallen in love with Psyche.
This has drawn the fury of his mother, Venus. The goddess then throws before Psyche a great mass of mixed crops.
There are $ n $ heaps of crops in total, numbered from $ 1 $ to $ n $.
Psyche needs to arrange them in a certain order, assume crops on the $ i $-th position is $ A_i $.
She is given some information about the final order of the crops:
1. the minimum value of $ A_1, A_2, ..., A_i $ is $ B_i $.
2. the maximum value of $ A_1, A_2, ..., A_i $ is $ C_i $.
She wants to know the number of valid permutations. As this number can be large, output it modulo $ 998244353 $.
Note that if there is no valid permutation, the answer is $ 0 $.
Input
The first line of input contains an integer $ T $ $ (1 \le T \le 15) $, which denotes the number of testcases.
For each test case, the first line of input contains single integer $ n $ $ (1 \le n \le 10 ^ 5) $.
The second line contains $ n $ integers, the $ i $-th integer denotes $ B_i $ $ (1 \le B_i \le n) $.
The third line contains $ n $ integers, the $ i $-th integer denotes $ C_i $ $ (1 \le C_i \le n) $.
Output
For each testcase, print the number of valid permutations modulo $ 998244353 $.
Sample Input
2 3 2 1 1 2 2 3 5 5 4 3 2 1 1 2 3 4 5
Sample Output
1 0
Hint
In the first example, there is only one valid permutation (2,1,3) . In the second example, it is obvious that there is no valid permutation.
题意:给出一排前i个值得最小值,给出一排前i个值得最大值,问有多少种满足情况的序列
思路:第一排非递增,第二排费递减,否则没有满足情况的序列。由刚才条件得可满足区间是递增的,但是题目要求每个值只能用一次,所以我们还得知道哪些值用过了。
细想可知,可满足区间内用过的值只会出现的前面,不用管后面。
所以①当出现新的最小值,可满足区间sum扩大(刚才的最小值minn[i-1]减去现在的最小值minn[i]减一)。
②当出现新的最大值,可满足区间sum扩大(现在的最大值t减去刚才的最大值maxx减一)。
③当最大值不变,这个数随机的,乘以可满足区间sum,然后sum--即可。
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