hdoj5791two【dp】

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 635    Accepted Submission(s): 289

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

 

题解就两个字水题然而对我来说却好难啊。。。

/* ***********************************************Author       : rycCreated Time : 2016-08-03 WednesdayFile Name    : E:\acmcode\hdoj\5791.cppLANGUAGE     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<stack>#include<map>#define MOD 1000000007llusing namespace std;const int maxn=1010;typedef long long LL;LL num1[maxn];LL num2[maxn];LL dp[maxn][maxn];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        for(int i=1;i<=n;++i){            scanf("%lld",&num1[i]);        }        for(int i=1;i<=m;++i){            scanf("%lld",&num2[i]);        }        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;++i){            for(int j=1;j<=m;++j){                dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1];                if(num1[i]==num2[j]){                    dp[i][j]+=dp[i-1][j-1]+1;                }                dp[i][j]%=MOD;            }        }        printf("%lld\n",dp[n][m]);    }    return 0;}