摆放着n堆石子。现要将石子有次序地合并成一堆，规定每次只能选相邻的2堆石子合并成新的一堆，并将新的一堆石子数记为该次合并的得分。求出将n堆石子合并成一堆的最小得分（或最大得分）合并方案。

动态规划。

代码用golang编写。代码如下：

package main

import (
    "fmt"
    "math"
)

func main() {
    arr := []int{1, 4, 2, 3}
    ret := StoneMerge(arr)
    fmt.Println(ret)

}
func sum(arr []int) []int {
    N := len(arr)
    s := make([]int, N+1)
    s[0] = 0
    for i := 0; i < N; i++ {
        s[i+1] = s[i] + arr[i]
    }
    return s
}
func w(s []int, l int, r int) int {
    return s[r+1] - s[l]
}
func StoneMerge(arr []int) int {
    if len(arr) < 2 {
        return 0
    }
    N := len(arr)
    s := sum(arr)
    dp := make([][]int, N)
    for i := 0; i < N; i++ {
        dp[i] = make([]int, N)
    }
    best := make([][]int, N)
    for i := 0; i < N; i++ {
        best[i] = make([]int, N)
    }
    for i := 0; i < N-1; i++ {
        best[i][i+1] = i
        dp[i][i+1] = w(s, i, i+1)
    }
    for L := N - 3; L >= 0; L-- {
        for R := L + 2; R < N; R++ {
            next := math.MaxInt64
            choose := -1
            for leftEnd := best[L][R-1]; leftEnd <= best[L+1][R]; leftEnd++ {
                cur := dp[L][leftEnd] + dp[leftEnd+1][R]
                if cur <= next {
                    next = cur
                    choose = leftEnd
                }
            }
            best[L][R] = choose
            dp[L][R] = next + w(s, L, R)
        }
    }
    return dp[0][N-1]
}

执行结果如下：