比较经典的dp了

可以用记忆化搜索，好写又快

f(i,j)
 if dp[i][j]!=-1 return dp[i][j]
 if i=0 and j=0 return dp[i][j]=1
 if i>0 and the ith item of s1 matches the i+jth item of s3
  if f(i-1,j) return dp[i][j]=1
 the same to j and s2
 return dp[i][j]=0