Two

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：

#include<bits/stdc++.h>
using namespace std;

const int pr=1000000007;

int dp[2010][2010];
int a[2010],b[2010];

int main(){
    int n,m;
    while(cin>>n>>m){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                dp[i][j]=0;
            }
        }
        for(int i=1;i<=n;i++){
            cin>>a[i];
        }
        for(int i=1;i<=m;i++){
            cin>>b[i];
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1];
                if(a[i]==b[j])
                dp[i][j]+=dp[i-1][j-1]+1;
                if(dp[i][j]<0)
                dp[i][j]+=pr;
                if(dp[i][j]>=pr)
                dp[i][j]%=pr;
            }
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}