HDU 5791 Two

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 163    Accepted Submission(s): 72

Problem Description

Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A' is a subsequence of A. B' is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.

 

Input

The input contains multiple test cases.

For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.

 

Output

For each test case, output the answer mod 1000000007.

 

Sample Input

3 2

1 2 3

2 1

3 2

1 2 3

1 2

 

Sample Output

2

3

 

Author

ZSTU

 状态方程 dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+(a[i]==b[j]?dp[i-1][j-1]+1:0)

可以这么理解:
比如序列

a :1 2

b :1 1 2

当i=2  j=3

首先不考虑a[2]和b[3]是什么  那么此时dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]把  i=2  j=3带入即可

再考虑 a[2]==b[3]时  这时 多增加了 1+dp[i-1][j-1]对，即a[2]和b[3]匹配和 a[2]b[3]和前面dp[i-1][j-1]匹配的对数。

a[2]!=b[3]时就不用考虑增加的....

 

注意 子序列可以不连续...一开始卡在这里了...zz

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/2 14:20:58
File Name     :p511.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 1000000007
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
ll dp[1100][1100];
int n,m;
int a[1100];
int b[1100];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    while(cin>>n>>m){
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int j=1;j<=m;j++)scanf("%d",&b[j]);
        cle(dp);
        ll x;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(a[i]==b[j])x=dp[i-1][j-1]+1;
                else x=0;
                dp[i][j]=(dp[i][j]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
                dp[i][j]=(dp[i][j]+x)%mod;
            }
        }
        printf("%I64d
",dp[n][m]);
    }
    return 0;
}