思路

通过动态规划计算出所有连续子序列的力量值之和，并将其存储在一个数组中然后排序，遍历一遍数组，找到相邻两个力量值之和的差的绝对值的最小值，然后输出这个答案就行了。

时间复杂度大概是 $O(n^2\log n)$。

来个 python 的代码

def min_power_diff(n, a):# 计算所有连续子序列的力量值之和
    prefix_sum = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix_sum[i] = prefix_sum[i - 1] + a[i - 1]
    # 计算差值并排序
    diffs = []
    for i in range(1, n + 1):
        for j in range(i, n + 1):
            diff = abs(prefix_sum[j] - prefix_sum[i - 1])
            diffs.append(diff)
    diffs.sort()
    # 找到相邻两个力量值之和的差的最小值
    min_diff = float('inf')
    for i in range(1, len(diffs)):
        min_diff = min(min_diff, diffs[i] - diffs[i - 1])
    return min_diff
n = int(input())
a = list(map(int, input().split()))
# 输出答案
result = min_power_diff(n, a)
print(result)