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`9`	`9`	但事实上本题里并不是整个区间统一倍增,而是间隔地倍增,怎么处理呢?由此我们发现,倍增的累加是间隔k发生的,故multiplier[i]只与multiplier[i-k]相关。由此我们知道更新的公式其实是`multiplier[i] = multiplier[i-k]*diff[i]`.
`10`	`10`
`11`		-特别注意,每一轮的multiplier需要再累加到一起。即`total_multiplier[i] = this_round_multiplier[i]`, 最终有`nums[i] = total_multiplier[i]`.
	`11`	+特别注意,每一轮的multiplier需要再累加到一起。即`total_multiplier[i] = multiplier[i]`, 最终有`nums[i] = total_multiplier[i]`.
`12`	`12`
`13`	`13`	`综上,对于k很小的时候,相同k的queries可以处理成同一组diff数组,然后根据diff更新multiplier数组。时间复杂度就是k*(q+n).`
`14`	`14`

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