hdu5791

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.

这题的大意是给你两个序列，让你求两个序列的公共子序列有多少个。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

long long dp[1100][1100];
const int mod = 1000000007;
int main(void)
{
    int n,m,i,j;
    int a[1100],b[1100];
    while(scanf("%d%d",&n,&m)==2)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=m;i++)
            scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                dp[i][j] = ((dp[i-1][j] + dp[i][j-1])%mod - dp[i-1][j-1] + mod)%mod;
                dp[i][j] %= mod;
                if(a[i] == b[j])
                    dp[i][j] = (dp[i][j]+ dp[i-1][j-1])%mod + 1;
                dp[i][j] %= mod;
            }
        cout << dp[n][m] << endl;
    }
}