Two

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5791

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

Hint

题意

给你两个数组，问你公共子序列的数量是多少

题解：

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]

如果a[i]=b[j]，dp[i][j]+=dp[i-1][j-1]+1

就好了

代码

#include <bits/stdc++.h>

using namespace std;

const int pr=1000000007;
int dp[2010][2010];
int n,m;
int a[2010],b[2010];

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        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++)
            scanf("%d",&a[i]);
        for(int i=1;i<=m;i++)
            scanf("%d",&b[i]);
     //   for(int i=1;i<=n;i++) dp[i][0]=1;
     //   for(int i=1;i<=m;i++) dp[0][i]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                dp[i][j]=dp[i-1][j]+dp[i][j-1]-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<<i<<" "<<j<<" "<<dp[i][j]<<endl;
            }
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}