[HDU 5791]Two

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1744    Accepted Submission(s): 775

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 21 2 32 13 21 2 31 2

 

Sample Output

23

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

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#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int n,m;#define N 1001int a[N],b[N];typedef long long ll;ll dp[N][N];#define mod 1000000007int main(){while(scanf("%d%d",&n,&m)==2){memset(dp,0,sizeof dp);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++){for(int j=1;j<=m;j++){if(a[i]==b[j])dp[i][j] = (dp[i-1][j]+dp[i][j-1]+1+mod)%mod;elsedp[i][j] = (dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;}}printf("%I64d\n",dp[n][m]%mod);}return 0;}

注意：

dp[i][j] = (dp[i-1][j]+dp[i][j-1]+1+mod)%mod;<pre name="code" class="cpp">dp[i][j] = (dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;

//因为有取模所以可能会减除负数。