1.SQLite解法：

SELECT g.job,(COUNT()+1)/2 AS start,COUNT()/2+1 AS end
FROM grade AS g
GROUP BY g.job
ORDER BY g.job

MySQL解法：

SELECT g.job,ceil(COUNT()1.0/2) AS start,floor(COUNT()1.0/2)+1 AS end
FROM grade AS g
GROUP BY g.job
ORDER BY g.job