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m个元素中选取n的组合

作者:互联网

目录

题目

有序排列

'''
输入,m,n, 输出0~m,中任选N的有序排列结果
例子:
10,3
->720 
4,2
->12
'''

$$
A_m^n=\frac{m!}{(m-n)!}\qquad(m>=n)
$$
$$
A_4^2=\frac{4!}{2!}=\frac{4\times3\times2\times1}{2\times1}=12
$$

def main(m,n):
    result=[]
    def handler(ldata,n,sel=[],cnt=0):
        if(cnt==n):
            result.append(sel)
        else:
            for item in ldata:
                handler([i for i in ldata if i!=item],n,sel=sel+[item],cnt=cnt+1)
        pass
        
    ldata=list(range(m))
    handler(ldata,n)
    return result
    pass
r=main(4,2)
for i in range(len(r)):
    print(f'{i+1:2d}  {r[i]}')
 1  [0, 1]
 2  [0, 2]
 3  [0, 3]
 4  [1, 0]
 5  [1, 2]
 6  [1, 3]
 7  [2, 0]
 8  [2, 1]
 9  [2, 3]
10  [3, 0]
11  [3, 1]
12  [3, 2]

无序组合

'''
输入,m,n, 输出0~m,中任选N的有序排列结果
例子:
10,3
->120 
4,2
->6
'''

$$
C_mn=\frac{A_mn}{n!}
=\frac{m!}{n!(m-n)!}\qquad(m>=n)
$$
$$
C_42=\frac{A_42}{2!}
=\frac{4!}{2!(4-2)!}=\frac{4\times3\times2\times1}{(2\times1)\times(2\times1)}=6
$$

def main(m,n):
    result={}
    def handler(ldata,n,sel=[],cnt=0):
        if(cnt==n):
            v=set(sel)
            result[tuple(v)]=None
        else:
            for item in ldata:
                handler([i for i in ldata if i!=item],n,sel=sel+[item],cnt=cnt+1)
        pass
        
    ldata=list(range(m))
    handler(ldata,n)
    return list(result.keys())
r=main(4,2)
for i in range(len(r)):
    print(f'{i+1:2d}  {r[i]}')
 1  (0, 1)
 2  (0, 2)
 3  (0, 3)
 4  (1, 2)
 5  (1, 3)
 6  (2, 3)

标签:ldata,cnt,frac,组合,元素,选取,handler,result,sel
来源: https://www.cnblogs.com/xiaoboz/p/14298238.html