当database是10个物种（A、B、C、E、F、G、H、J、I、K）时，进行all vs all 比对结果是：
此时reference是物种A的第一个基因：即用10个物种的genome中的所有基因比对物种A的基因1
eg：如下
$ cat A_R00001_A.fa.70.blastp.m1|grep '1_0' -A 2 > r.sh
Query= A_R00001_A
(423 letters)
Database: all.10
165,229 sequences; 72,505,122 total letters
Searching..................................................done
Score E
Sequences producing significant alignments: (bits) Value
A_R00001_A 855 0.0
B_R00001_B 766 0.0
1_0 1 PITLWLFYTHWLPGSPTTHHRVQGTTAMKDMVKCRVLKAHPCQECGKSFSKKGNLKRHQR 60
A_R00001_A 1 ............................................................ 60
B_R00001_B 4 ..l.......w..dAA....L..rv......................... 53
--
...
1_0 361 GKCFPHKRHLIKHQLLHSRGGAYKCGVCGKRYRLKKYLRRHQKIHMREGTAPCSKPGYTT 420
A_R00001_A 361 ............................................................ 420
B_R00001_B 357 ..........................d..................t.........wee.. 416
--
1_0 421 QTA 423
A_R00001_A 421 ... 423
B_R00001_B 417 R.G 419
1.Sequences producing significant alignments:大致情况，并计算score值，E值
关于score值：
he bit score gives an indication of how good the alignment is; the higher the score, the better the alignment. In general terms, this score is calculated from a formula that takes into account the alignment of similar or identical residues, as well as any gaps introduced to align the sequences.
2.具体情况，
1_0：表示reference的的第一个基因，即A物种的基因1
B_R00001_B：表示物种B的基因1与该reference（A物种的基因1） 的比对情况，从第4个AA开始到第53个AA结束
..l.......w..dAA....L..rv........................长度为53-4+1=50个；.是比对成功；空白是B物种基因1没有序列；60个为一组
如果物种A有10个基因，则将会把10个基因全部比对一遍：
$cat A_R00001_A.fa.70.blastp.m1|grep 'Query' >> r.sh Query= A_R00001_A Query= A_R00002_A Query= A_R00003_A Query= A_R00004_A Query= A_R00005_A Query= A_R00006_A Query= A_R00007_A Query= A_R00008_A Query= A_R00009_A Query= A_R00010_A