square subsequence, now 12th

I think I had squared subsequence and I did. Now placed at 12th!

The idea is to try all possible ways to break s[] into s1[] and s2[] and count the number of equal subseq in both s1[] and s2[] with the last char of s1 matched.

The recurrence is common[n1][n2] = common[n1][n2-1] + common[n1-1][n2] - common[n1-1][n2-1]
we add back common[n1-1][n2-1] if s1[n1-1] match s2[n2-1].

# Status Signal Time
1 Passed Your code produced correct output for this testcase. 0.07061
2 Passed Your code produced correct output for this testcase. 0.070552
3 Passed Your code produced correct output for this testcase. 0.564724
4 Passed Your code produced correct output for this testcase. 0.503675
5 Passed Your code produced correct output for this testcase. 0.383056
6 Passed Your code produced correct output for this testcase. 0.514264
7 Passed Your code produced correct output for this testcase. 0.695142
8 Passed Your code produced correct output for this testcase. 0.524266
9 Passed Your code produced correct output for this testcase. 0.556575
10 Passed Your code produced correct output for this testcase. 0.473493

Rank Hacker Country Score Submissions Solved
11 diogoaos Brazil 811.00 56 21
12 lantimilan USA 784.00 79 19
13 liangjiaxing Canada 778.00 50 19
14 jin.ubc Canada 763.00 42 20
15 uwi Japan 760.00 39 19
15 megaterik Belarus 760.00 102 22
17 hadi Iran 740.00 48 19
18 Amtrix Germany 735.00 20 19
19 chhung6 Hong Kong 732.00 76 19
20 ttim Kazakhstan 730.00 95 19