HDU 6078 dp

题目

Wavel Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 644    Accepted Submission(s): 335

Problem Description

Have you ever seen the wave? It's a wonderful view of nature. Little Q is attracted to such wonderful thing, he even likes everything that looks like wave. Formally, he defines a sequence a1,a2,...,an as ''wavel'' if and only if a1<a2>a3<a4>a5<a6...



Picture from Wikimedia Commons


Now given two sequences a1,a2,...,an and b1,b2,...,bm, Little Q wants to find two sequences f1,f2,...,fk(1≤fi≤n,fi<fi+1) and g1,g2,...,gk(1≤gi≤m,gi<gi+1), where afi=bgi always holds and sequence af1,af2,...,afk is ''wavel''.

Moreover, Little Q is wondering how many such two sequences f and g he can find. Please write a program to help him figure out the answer.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 2 integers n,m(1≤n,m≤2000) in the first line, denoting the length of a and b.

In the next line, there are n integers a1,a2,...,an(1≤ai≤2000), denoting the sequence a.

Then in the next line, there are m integers b1,b2,...,bm(1≤bi≤2000), denoting the sequence b.

 

Output

For each test case, print a single line containing an integer, denoting the answer. Since the answer may be very large, please print the answer modulo 998244353.

 

Sample Input

13 51 5 34 1 1 5 3

 

Sample Output

10
Hint
(1)f=(1),g=(2).(2)f=(1),g=(3).(3)f=(2),g=(4).(4)f=(3),g=(5).(5)f=(1,2),g=(2,4).(6)f=(1,2),g=(3,4).(7)f=(1,3),g=(2,5).(8)f=(1,3),g=(3,5).(9)f=(1,2,3),g=(2,4,5).(10)f=(1,2,3),g=(3,4,5).

题目大意

 看hint应该很好理解

解题思路

有点求最长公共子序列的意思，通过DP求解该点之前的波峰和波谷的数量。有机会再写一次吧

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define LL long long#define read(a) scanf("%d",&a)const int mod=998244353;int a[2000+10];int b[2000+10];LL dp[2000+10][2];LL sum[2000+10][2];void work(){    memset(dp,0,sizeof(dp));    memset(sum,0,sizeof(sum));    int n,m;    read(n),read(m);    LL ans=0;    for(int i=1;i<=n;i++)        read(a[i]);    for(int i=1;i<=m;i++)        read(b[i]);    for(int i=1;i<=n;i++)    {        LL peak=1;        LL val=0;        for(int j=1;j<=m;j++)        {            if(a[i]==b[j])            {                dp[j][0]=peak;                dp[j][1]=val;                ans=(ans+(peak+val)%mod)%mod;                sum[j][0]=(sum[j][0]+dp[j][0])%mod;                sum[j][1]=(sum[j][1]+dp[j][1])%mod;            }            else if(a[i]>b[j])            {                val=(val+sum[j][0])%mod;            }            else            {                peak=(peak+sum[j][1])%mod;            }        }    }    printf("%lld\n",ans);}int main(){    int T;    read(T);    while(T--)    {        work();    }    return 0;}