喜欢切数组的红

k = int(input())
nums = list(map(int, input().split()))


def count_solution(nums_list):
    sum_list = []
    positive_list = []
    sum_temp = 0
    for num in nums_list:
        if num > 0:
            positive_list.append(1)
        else:
            positive_list.append(0)
        sum_temp += num
        sum_list.append(sum_temp)
    first_son_sum = sum_list[-1] / 3
    # 寻找第一次分割
    frist_maybe_c = sum_list.count(first_son_sum)
    if frist_maybe_c == 0:
        return 0
    else:
        frist_maybe_index = []
        i = -1
        while len(frist_maybe_index) < frist_maybe_c:
            i = sum_list.index(first_son_sum, i + 1)
            frist_maybe_index.append(i)
        # 寻找第二次分割位置
        second_maybe_index = []
        second_son_sum = 2 * first_son_sum
        second_maybe_c = sum_list.count(second_son_sum)
        i = 0  # 第二次不可能从第一位开始分割
        while len(second_maybe_index) < second_maybe_c:
            i = sum_list.index(second_son_sum, i + 1)
            # 防止出现和为零时导致第二次分割的位置在末尾
            if i < len(nums_list):
                second_maybe_index.append(i)
        all_solotion = []
        for m in frist_maybe_index:
            for n in second_maybe_index:
                if m < n:  # 呃附着的出现可能会导致N大于M
                    # 保证三个数组都有正shu存在
                    if (
                        sum(positive_list[: m + 1]) > 0
                        and sum(positive_list[m : n + 1]) > 0
                        and sum(positive_list[n:]) > 0
                    ):
                        all_solotion.append([m, n])
    return len(all_solotion)


print(count_solution(nums))
#超时，通过率50%，就当给大家提供思路吧