hdu 5719 Arrange（排列组合）

Arrange

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 21    Accepted Submission(s): 9

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  Ai.

She is given some information about the final order of the crops:

1. the minimum value of  A1,A2,...,Ai is  Bi.

2. the maximum value of  A1,A2,...,Ai is  Ci.

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≤T≤15), which denotes the number of testcases.

For each test case, the first line of input contains single integer  n  (1≤n≤105).

The second line contains  n integers, the  i-th integer denotes  Bi  (1≤Bi≤n).

The third line contains  n integers, the  i-th integer denotes  Ci  (1≤Ci≤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：Ai是1~n的某种排列，Bi表示前i个数的最小值，Ci表示前i个数的最大值

先给出B和C，求A序列有多少种可能

思路：如果Bi>B(i-1),CiCi,B1!=C1那么都是不合法的

然后我们枚举i点

1:Bi==Bi-1&&Ci==Ci-1  说明Ai是A1~A（i-1）中出现过的一个数，一共有max-min+1种可能性，不过因为最后的序列必须包含1~n，所以这里设置一个计数器，表示从min-max中有多少个数是已经存在的了，这里必须填一个不存在的数。结果用乘法原理乘起来就OK

2:Bi

3::Ci>Ci-1&&Bi==Bi-1 说明此时Ai==Ci

4:BiCi-1  Ai又要小于当前最小值又要大于当前最大值，显然矛盾，直接输出0

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 200050
#define mod 998244353
int b[N],c[N];
int ans[N];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&b[i]);
        for(int i=1; i<=n; i++)
            scanf("%d",&c[i]);
        int flag=0;
        for(int i=2; i<=n; i++)
            if(b[i]>c[i]||b[i]>b[i-1]||c[i]n||c[i]<1||c[i]>n)
            {
                flag=1;
                break;
            }
        if(b[1]!=c[1]||flag||b[1]<1||b[1]>n)
        {
            printf("0\n");
            continue;
        }
        __int64 ans=1;
        __int64 minn=b[1],maxn=b[1];
        int num=1;
        for(int i=2; i<=n; i++)
        {
            if(b[i]==b[i-1]&&c[i]==c[i-1])
                {
                ans=ans*(maxn-minn-num+1)%mod;
                num++;
                }
            else if(b[i]c[i-1])
                {
                    maxn=c[i];
                    num++;
                }
            else
            {
                flag=1;
                break;
            }
        }
        if(flag) printf("0\n");
        else printf("%I64d\n",ans);
    }
    return 0;
}