hdu 5791 (DP) Two

hdu 5791

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1421    Accepted Submission(s): 630

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

 

求两个序列的相同的公共子序列的个数，取模。

DP[i][j]表示到a数组的第 i 个和b数组的第 j 个之间的个数，可以归纳出，当a[i]==b[j]时，dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;

当不等的时候，dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1].(仔细琢磨琢磨，会发现很有道理)因为有减法的运算所以最后答案可能出现

负数，注意处理一下。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int mod = 1000000007;
ll dp[1010][1010];
int a[1010],b[1010];

int main()
{
    int n,m;
    while (~scanf("%d%d",&n,&m)){
        for (int i=1 ; i<=n ; i++) scanf("%d",&a[i]);
        for (int i=1 ; i<=m ; i++) scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        for (int i=1 ; i<=n ; i++){
            for (int j=1 ; j<=m ; j++){
                if (a[i]==b[j]) dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;
                else dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
                dp[i][j]%=mod;
            }
        }
        printf("%I64d\n",(dp[n][m]+mod)%mod);
    }
    return 0;
}