sched/numa: Fix a possible divide-by-zero

sched_clock_cpu() may not be consistent between CPUs. If a task
migrates to another CPU, then se.exec_start is set to that CPU's
rq_clock_task() by update_stats_curr_start(). Specifically, the new
value might be before the old value due to clock skew.

So then if in numa_get_avg_runtime() the expression:

  'now - p->last_task_numa_placement'

ends up as -1, then the divider '*period + 1' in task_numa_placement()
is 0 and things go bang. Similar to update_curr(), check if time goes
backwards to avoid this.

[ peterz: Wrote new changelog. ]
[ mingo: Tweaked the code comment. ]
Signed-off-by: Xie XiuQi <xiexiuqi@huawei.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Cc: cj.chengjian@huawei.com
Cc: <stable@vger.kernel.org>
Link: http://lkml.kernel.org/r/20190425080016.GX11158@hirez.programming.kicks-ass.netSigned-off-by: Ingo Molnar <mingo@kernel.org>

kernel/sched/fair.c

@@ -2007,6 +2007,10 @@ static u64 numa_get_avg_runtime(struct task_struct *p, u64 *period)
 	if (p->last_task_numa_placement) {
 		delta = runtime - p->last_sum_exec_runtime;
 		*period = now - p->last_task_numa_placement;
+
+		/* Avoid time going backwards, prevent potential divide error: */
+		if (unlikely((s64)*period < 0))
+			*period = 0;
 	} else {
 		delta = p->se.avg.load_sum;
 		*period = LOAD_AVG_MAX;