HDU_5791_Two（简单dp）

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1051    Accepted Submission(s): 489

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

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

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题意

给你两个数串

问你两个数串有多少子串一致

子串不一定是连续的

解题思路

首先这个问题要想到是个dp问题

dp[i][j]代表第一个串到I位第二个串到j位相同子串的个数

则dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[I-1][j-1]减掉多的部分

但是末尾如果匹配的话

那么只要加上都减掉末尾的匹配数量再加1就可以了

因为这些子串都可以加上末尾这个元素构成新的匹配，而末尾的匹配自身也构成1个答案

#include 
#include 
#include 
#include 
using namespace std;

typedef unsigned long long LL;
const int M=1005;
const int MO=1e9+7;
LL dp[M][M];
//dp[i][j] 第一个串到i位置第二个串到j位置相同的前缀个数

int nu1[M],nu2[M];

int main()
{
    //freopen("1.in","r",stdin);
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&nu1[i]);
        for(int i=1;i<=m;i++)
            scanf("%d",&nu2[i]);
        memset(dp,0,sizeof(dp));
        //for(int i=1;i