HDU 5791 Two

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2692    Accepted Submission(s): 1156

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

题意：给定两个数列。求出它们公共子序列的数量并对1e9+7取模

一开始没什么思路，后面看了大佬们的代码发现可以用dp做，emmm

切入正题，我们可以用dp[i][j]表示A序列到i位，B序列到j位相同子序列的个数

那么a[i]==b[j]的话

dp[i][j]=(dp[i-1][j]+d[i][j-1]+1)%mod;

不满足的话(这里要考虑防止相减得出负数的情况，所以加上mod）

dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;

附上全部代码：

#include <iostream>
#include <cstdio>
#define mod 1000000007
using namespace std;
int a[1005],b[1005];
long long dp[1005][1005];
int main(){
    int n,m;
//    freopen("in.in","r",stdin);
    while(cin>>n>>m){
        dp[0][0]=dp[0][1]=dp[1][0]=0;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        for(int i=1;i<=m;i++)
            cin>>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;
                }
                else{
                    dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
                }
            }
        }
        cout<<dp[n][m]<<endl;
    }
}