题目链接：点击打开链接
题目大意：略。
解题思路：
- 解决方案(1)：自变量，算出每一行记录与上一行记录的累加即可
解决方案(2)：t1.event_date>=t2.event_date + GROUP BY t1.player_id,t1.event_date 是关键：（自连接）用 t1 比自己还小的所有在 t1.player_id,t1.event_date 范围内的时间，就是到达 t1 时间的总和游戏数
AC 代码
--解决方案(1) SELECTRS.player_id, RS.event_date, RS.games_played_so_farFROM (SELECTplayer_id, event_date, CONVERT(:=IF(<>player_id, games_played, +games_played), SIGNED) games_played_so_far, :=player_idFROMActivity, (SELECT:=0, :=-1) initORDERBYplayer_id, event_date) RS; --解决方案(2) SELECTt1.player_id, t1.event_date, SUM(t2.games_played) games_played_so_farFROMActivityt1,Activityt2WHEREt1.player_id=t2.player_idANDt1.event_date>=t2.event_dateGROUPBYt1.player_id,t1.event_date;
