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题目链接：

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)SELECT RS.player_id, RS.event_date, RS.games_played_so_farFROM (SELECT player_id, event_date, CONVERT(@sum:=IF(@p<>player_id, games_played, @sum + games_played), SIGNED) games_played_so_far, @p:=player_idFROM Activity, (SELECT @sum:=0, @p:=-1) initORDER BY player_id, event_date) RS;-- 解决方案(2)SELECT t1.player_id,       t1.event_date,       SUM(t2.games_played) games_played_so_farFROM Activity t1,Activity t2WHERE t1.player_id=t2.player_id AND t1.event_date>=t2.event_dateGROUP BY t1.player_id,t1.event_date;

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