HDU 5791 Two

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(1N,M1000). 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

开个树状数组乱搞一通就过了

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
using namespace std;
typedef __int64 LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e3+ 10;
int T, n, m, dp[N][N], a[N], b[N];
int f[N][N];

int get(int x, int y)
{
    int res = 0;
    for (int i = x; i; i -= low(i))
    {
        for (int j = y; j; j -= low(j))
        {
            (res += f[i][j]) %= mod;
        }
    }
    return res;
}

void insert(int x, int y, int z)
{
    for (int i = x; i <= n; i += low(i))
    {
        for (int j = y; j <= m; j += low(j))
        {
            f[i][j] = (f[i][j] + z) % mod;
        }
    }
}

int main()
{
    //scanf("%d", &T);
    //while (T--)
    while (scanf("%d%d", &n, &m) != EOF)
    {
        rep(i, 1, n) scanf("%d", &a[i]);
        rep(i, 1, m) scanf("%d", &b[i]);
        rep(i, 1, n) rep(j, 1, m) f[i][j] = 0;
        int ans = 0;
        rep(i, 1, n)
        {
            rep(j, 1, m)
            {
                if (a[i] == b[j])
                {
                    dp[i][j] = (1 + get(i - 1, j - 1)) % mod;
                    insert(i, j, dp[i][j]);
                    ans = (ans + dp[i][j]) % mod;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}


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