#### 原题说明

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose any two rocks and smash them together.  Suppose the stones have weights `x` and `y` with `x <= y`.  The result of this smash is:

• If `x == y`, both stones are totally destroyed;
• If `x != y`, the stone of weight `x` is totally destroyed, and the stone of weight `y` has new weight `y-x`.

At the end, there is at most 1 stone left.  Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.)

Example 1:

`Input: [2,7,4,1,8,1]Output: 1Explanation: We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,we can combine 1 and 1 to get 0 so the array converts to [1] then that’s the optimal value. `

Note:

1. `1 <= stones.length <= 30`
2. `1 <= stones[i] <= 100`

#### 解题思路

`S1 + S2 = total_weight`
`if S1 <= S2 => S2 - S1 = total_weight - S1 - S1 = total_weight - 2 * S1`

`dp[i]``true`如果我们可以使得一组`stones`的重量之和`S1`等于`i`，反之则为`false`

#### 归纳总结

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