hdu 5719 Arrange 贪心

Arrange

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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.

思路：最小值最大值成递减递增函数；前面取的数肯定在之后的区间内，flag一下；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e5+10,M=1e7+10,inf=1e9+10;
const ll mod=998244353;
int a[N];
int b[N];
int main()
{
    int x,y,z,i,t;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&x);
        ll ans=1;
        int flag=1;
        for(i=0;i<x;i++)
        scanf("%d",&a[i]);
        for(i=0;i<x;i++)
        scanf("%d",&b[i]);
        if(a[0]!=b[0])
        ans=0;
        for(i=1;i<x;i++)
        {
            if(a[i]!=a[i-1]&&b[i]!=b[i-1])
            ans=0;
            else if(a[i]>a[i-1])
            ans=0;
            else if(b[i]<b[i-1])
            ans=0;
            else if(a[i]!=a[i-1]||b[i]!=b[i-1])
            flag++;
            else
            {
                ans*=(b[i]-flag-a[i]+1);
                flag++;
                ans%=mod;
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/jhz033/p/5679646.html