BestCoders, bestcoderhdu

BestCoders, bestcoderhdu

Senior's String Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 466 Accepted Submission (s): 183


Problem DescriptionXuejiejie loves strings most. In order to win the favor of her, a young man has two strings X , Y To Xuejiejie. Xuejiejie has never seen such beautiful strings! These days, she is very happy. but Xuejiejie is missish so much, in order to cover up her happiness, she asks the young man a question. in face of Xuejiejie, the young man is flustered. so he asks you for help.

The question is that:
Define L As the length of the longest common subsequence X And Y . (The subsequence does not need to be continuous
In the string, and a string of length L Has 2L Subsequences containing the empty string). Now Xuejiejie comes up with all subsequences of length L Of string X , She wants to know the number of subsequences which is also the subsequence of string Y .
 
InputIn the first line there is an integer T , Indicates the number of test cases.

In each case:

The first line contains string X , A non-empty string consists of lowercase English letters.

The second line contains string Y , A non-empty string consists of lowercase English letters.

1 ≤ | X |, | Y | ≤ 1000 , | X | Means the length X .
OutputFor each test case, output one integer which means the number of subsequences of length L Of X Which also is the subsequence of string Y Modulo 109 + 7 .
Sample Input

2abaaab

 
Sample Output

12

 
SourceBestCoder Round #47 ($)
Recommendhujie

Note: DP d (I, j) indicates the maximum length of common substrings. f (I, j) indicates the number of substrings with the length of d (I, j) in Y.

Recursive Formula: It is very simple. Select a condition and analyze the situation.

Here we choose not to take X (I)

I. if d (I, j) = d (I-1, j) then f (I, j) + = f (I-1, j );

II. if d (I, j) = d (I-1, P-1) + 1 | X (I) = Y (p) & p <= j then f (I, j) + = f (I-1, P-1 ).

Node: Initial Conditions -- d (0, I) = d (I, 0) = 0, f (0, I) = f (0, I) = 1, f (I, j) = 0;

<pre name="code" class="cpp">#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define maxn 1005#define MOD 1000000007char s1[maxn], s2[maxn];int d[maxn][maxn];int f[maxn][maxn];int pos[maxn][27];void solve(){    int len1 = strlen(s1+1);    int len2 = strlen(s2+1);    memset(f, 0, sizeof(f));    for(int i=0; i<=len2; i++)    {        d[0][i] = d[i][0] = 0;        f[0][i] = f[i][0] = 1;    }    for(int i=1; i<=len1; i++)        for(int j=1; j<=len2; j++)            if(s1[i] == s2[j])                d[i][j] = d[i-1][j-1] + 1;            else                d[i][j] = max(d[i-1][j], d[i][j-1]);    memset(pos, 0, sizeof(pos));    for(int i=1; i<=len2; i++)    {        for(int j=0; j<26; j++)            pos[i][j] = pos[i-1][j];        pos[i][s2[i]-'a'] = i;    }    f[0][0] = 1;    for(int i=1; i<=len1; i++)        for(int j=1; j<=len2; j++)        {            if(d[i][j] == 0)            {                f[i][j] = 1;                continue;            }            if(d[i][j] == d[i-1][j])                f[i][j] = (f[i][j] + f[i-1][j]) % MOD;            int p = pos[j][s1[i]-'a'];            if(p && d[i][j] == d[i-1][p-1] + 1)                f[i][j] = (f[i][j] + f[i-1][p-1]) % MOD;        }        printf("%d\n", f[len1][len2]);}int main(){    //freopen("input.txt", "r", stdin);    //freopen("output.txt", "w", stdout);    int T;    cin>>T;    while(T--)    {        scanf("%s", s1+1);        scanf("%s", s2+1);        solve();    }    return 0;}

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