Two HDU

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)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
dp题，原型是最长公共子序列，只不过这个要求个数。
状态：

dp[x][y]:a序列1到x和b序列1到y符合要求的答案

状态转移方程：

当a[x]=b[y]，则dp[x][y]=dp[x][y−1]+dp[x−1][y]+1

当不等，dp[x][y]=dp[x][y−1]+dp[x−1][y]−dp[x−1][y−1]

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define mod 1000000007using namespace std;int n,m;int dp[1003][1003];int a[1003],b[1003];int solve(int x,int y){    if(dp[x][y]!=-1)        return dp[x][y];    if(a[x]==b[y])        return dp[x][y]=(solve(x,y-1)+solve(x-1,y)+1)%mod;    int temp=solve(x,y-1)+solve(x-1,y)-solve(x-1,y-1);    if(temp<0)        return dp[x][y]=(temp+mod)%mod;    else        return dp[x][y]=temp%mod;}int main(){    while(scanf("%d%d",&n,&m)==2)    {        for(int i=1;i<=n;i++)            scanf("%d",a+i);        for(int i=1;i<=m;i++)            scanf("%d",b+i);        memset(dp,-1,sizeof(dp));            for(int i=0;i<=1000;i++)        dp[i][0]=0,dp[0][i]=0;        printf("%d\n",solve(n,m));    }    return 0;}