尝试将jQuery鼠标悬停转换为页面加载

时间:2018-05-13 22:59:41

标签: javascript jquery

此代码目前仅在鼠标悬停和鼠标移出事件期间加载。我想在页面加载时执行。我已经尝试修改事件以阅读...但它只是停止一起运行。有什么建议吗?另外,有关如何添加复选框以禁用该功能的任何建议吗?

global a



a = list(map(int , input().split()))
global DP
global neg_
global sum_seq
sum_seq = [0]*len(a)


def solve(i):
    k = 1
    a1 = []
    b1 = []
    while i + 1:
        if i == len(a) - 1 or i == len(a) - 2:
            if i == len(a) - 1:
                a1.append(a[i])
                sum_seq[i] = a[i]
            if i == len(a) - 2:
                if a[i] > sum_seq[i + 1]:
                    b1.append(a[i])
                    sum_seq[i] = a[i]
                else:
                    b1.append(a[i+1])
                    sum_seq[i] = sum_seq[i+1]
        else:
            if k % 2:
                if sum_seq[i + 1] < sum_seq[i+2] + a[i]:
                    a1.append(a[i])
                    sum_seq[i] = sum_seq[i+2] + a[i]
                    k += 1
                else:
                    sum_seq[i] = sum_seq[i+1]
                    a1.clear()
                    a1 = b1[:]
                    k += 1  
            else:
                if sum_seq[i+1] < sum_seq[i+2] + a[i]:
                    b1.append(a[i])
                    sum_seq[i] = sum_seq[i+2] + a[i]
                    k += 1
                else:
                    sum_seq[i] = sum_seq[i+1]
                    b1.clear()
                    b1 = a1[:]
                    k += 1
        i -= 1
    print(*a1)
    print(*b1)
    print(sum(a1))
    print(sum(b1))
solve(len(a) - 1)

1 个答案:

答案 0 :(得分:0)

使用以下内容:

jQuery(window).on("load", function(){
  sparkle.over();
});