银币次品求解问题---二分法和三分法比较（python代码）

N块银币中有一块不合格，已知不合格的银币比正常银币重。现有一台天平，请利用它找出不合格的银币，并且要求使用天平的次数最少。

def sum(left,right):
    num=0
    for i in range(left,right+1):
        num=list1[i]+num

    return num

def search(left,right):
    mid=(left+right)//2
    global f
    if left==right-1:

        f = f + 1
        if list1[left]>list1[right]:

            return left
        else:return right

    if (right-left)%2==0:#说明是奇数
        f = f + 1
        if sum(left,mid-1)>sum(mid+1,right):


            return search(left,mid-1)
        elif sum(left,mid-1)<sum(mid+1,right):


            return search(mid+1,right)
        else :return mid

    else :#说明是偶数
        f = f + 1
        if sum(left,mid)>sum(mid+1,right):

            return search(left,mid)
        else :

            return search(mid+1,right)

if __name__ == '__main__':
    N = int(input("请输入银币总数："))
    s = int(input("请输入不合格银币所在位置："))
    left = 1
    right = N
    f = 0

    list1 = [1 for i in range(N + 1)]
    list1[s] = 2
    flag=search(1,N)

    print(f"不合格银币位置：{flag}")
    print(f"比较次数：{f}")

下面是使用三分法的代码，最后比较一下使用二分和三分的优缺点。

def sum(left,right):
    num=0
    for i in range(left,right+1):
        num=list1[i]+num

    return num

def search(left,right):
    mid=(right-left+1)//3
    k=(right-left+1)%3
    global f

    if left==right:
        f=f+1
        return left
    elif left==right-1:
        f=f+1
        if list1[left]>list1[right]:
            return left
        else :return right

    f = f + 1
    if sum(left,left+mid-1)>sum(left+mid,left+mid*2-1):

        return search(left,left+mid-1)
    elif sum(left,left+mid-1)<sum(left+mid,left+mid*2-1):

        return search(left+mid,left+mid*2-1)
    elif sum(left+mid*2,left+mid*3-1)>sum(left,left+mid-1):

        return search(left+mid*2,left+mid*3-1)
    else:
        if k==1:
            return right
        elif k==2:
            f=f+1
            if list1[right-1]>list1[right]:
                return right-1
            else :return right













if __name__ == '__main__':
    N = int(input("请输入银币总数："))
    s = int(input("请输入不合格银币所在位置："))
    left = 1
    right = N
    f = 0

    list1 = [1 for i in range(N + 1)]
    list1[s] = 2
    flag=search(1,N)

    print(f"不合格银币位置：{flag}")
    print(f"比较次数：{f}")

粗略的发现，三分法在银币总数大时的比较次数更少，更适合。