Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1533    Accepted Submission(s): 676

 

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

 

 

题意：

求公共子序列数量。

 

dp[i][j]表示第一个串考虑到i位，第二个串考虑到j位的答案是多少。

那么dp[i][j] = dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1] ，需要特别判断a[i]=b[j]时，dp[i][j]+=dp[i-1][j-1]+1。

 

附AC代码：

1 #include
     
       2 using namespace std; 3  4 const int pr=1000000007; 5  6 int dp[2010][2010]; 7 int a[2010],b[2010]; 8  9 int main(){10     int n,m;11     while(cin>>n>>m){12         for(int i=1;i<=n;i++){13             for(int j=1;j<=m;j++){14                 dp[i][j]=0;15             }16         }17         for(int i=1;i<=n;i++){18             cin>>a[i];19         }20         for(int i=1;i<=m;i++){21             cin>>b[i];22         }23         for(int i=1;i<=n;i++){24             for(int j=1;j<=m;j++){25                 dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1];26                 if(a[i]==b[j])27                 dp[i][j]+=dp[i-1][j-1]+1;28                 if(dp[i][j]<0)29                 dp[i][j]+=pr;30                 if(dp[i][j]>=pr)31                 dp[i][j]%=pr;32             }33         }34         cout<
      
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