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Dynamic_Programming/3444.Minimum-Increments-for-Target-Multiples-in-an-Array
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`7`	`7`	2. nums[i]是state里某些元素的(公)倍数。于是我们需要枚举state里的子集subset,就有`dp[i][state]=dp[i-1][state-subset]+cost(nums[i], targets[subset]`。这里的cost,显然就是将nums[i]增加至targets[subset]所对应的最小公倍数即可。
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`9`	`9`	`最终答案就是dp[n][(1<<m)-1],表示将nums的所有元素考察完之后,实现覆盖了targets里所有元素的倍数,所对应的最小代价。`
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`11`		`-有个细节就是计算targets[subset]的最小公倍数时可能会整型溢出,可以将这部分跳过。这是因为targets的元素不超过1e4,那么我们所需要操作的代价最多也只需要4*1e4,就可以保证在nums里构造出targets里每个元素的倍数。`

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