数组分组

import itertools
n = int(input())
A = list(map(int, input().split()))
a = []
b = []
rest = []
for i in A:
    if i % 5 == 0:
        a.append(i)
    else:
        if i % 3 == 0:
            b.append(i)
        else:
            rest.append(i)
if rest == []:
    if sum(a) == sum(b):
        print('true')
    else:
        print('false')
else:
    for i in range(1, int(len(rest)/2)+2):
        for combin in itertools.combinations(rest, i):
            if sum(a)+sum(list(combin)) == sum(b)+sum(rest)-sum(list(combin))&nbs***bsp;sum(b)+sum(list(combin)) == sum(a)+sum(rest)-sum(list(combin)):
                print('true')
                exit()
    print('false')

def dfs(five,three,other):
    if not other:
        if sum(five) == sum(three):
            return True
        else:
            return False
    else:
        return dfs(five+other[:1],three,other[1:])&nbs***bsp;dfs(five,three+other[:1],other[1:])

n = int(input())
s = list(map(int,input().split()))
a,b,temp = [],[],[]
for i in s:
    if i % 5 == 0:
        a.append(i)
    elif i % 3 == 0:
        b.append(i)
    else:
        temp.append(i)
print('true' if dfs(a,b,temp) else 'false')

import sys
from itertools import combinations

while True:
    try:
        n = input()
        x = list(map(int,input().split()))
        a3 =[0] + [i for i in x if i%3==0 and i%5!=0]
        a5 = [0] + [i for i in x if i%5==0]
        a0 = [0] + [i for i in x if i not in a3+a5]
        k = 'false'
        # print(a3,a5,a0,sep = '\n')
        for i in range(1,len(a0)+1):
            if k=='true':
                break
                    
            for j in combinations(a0,i):
                if (sum(a3)+sum(a0)-sum(a5))%2==0 and sum(j) == (sum(a3)+sum(a0)-sum(a5))//2:
                    k = 'true'
                    # print(j,sum(j),(sum(a3)+sum(a0)-sum(a5)))
                    break
        print(k)
 
    except:
        break

import itertools


def arr_grouping(arr):
    # 分组：能被5整除、能被3整除、其余的元素
    five_multiple = [x for x in arr if x % 5 == 0]
    three_multiple = [x for x in arr if x % 3 == 0 and x not in five_multiple]
    arr_surplus = [x for x in arr if x not in five_multiple and x not in three_multiple]

    # 总和计算
    sum_five = sum(five_multiple)
    sum_three = sum(three_multiple)

    # 尝试将 arr_surplus 组合成与剩余元素相等的总和
    total_sum = sum(arr_surplus)
    target_sum = (total_sum + sum_five - sum_three) / 2

    # 如果目标和不是整数，表示不能均分
    if target_sum != int(target_sum):
        return False

    target_sum = int(target_sum)

    # 检查 arr_surplus 的组合是否能达到 target_sum
    for k in range(len(arr_surplus) + 1):
        # 使用组合来检测目标和
        for comb in itertools.combinations(arr_surplus, k):
            if sum(comb) == target_sum:  # sum([]) = 0
                return True
    return False


# 输入读取
n = int(input())
arr = [int(x) for x in input().split()]

# 输出结果
print(str(arr_grouping(arr)).lower())

def fun(sum_3, sum_5, nums_other):
    if len(nums_other) == 0:
        if sum_3 == sum_5:
            return True
        else:
            return False
    else:
        fun1 = fun(sum_3 + nums_other[0], sum_5, nums_other[1::])
        fun2 = fun(sum_3, sum_5 + nums_other[0], nums_other[1::])
        return fun1 or fun2


n = int(input())
nums = list(map(int, input().split()))
nums_5 = []
nums_3 = []
nums_other = []
for i in nums:
    if i % 5 == 0:
        nums_5.append(i)
    elif i % 3 == 0:
        nums_3.append(i)
    else:
        nums_other.append(i)
sum_3 = sum(nums_3[::])
sum_5 = sum(nums_5[::])

print('true') if fun(sum_3, sum_5, nums_other) else print('false')

def add(a5,a3,a0):#看别人代码写的
    if len(a0) == 0:
        if a5 == a3:
            return True
        else:
            return False
    else:
        return add(a5+a0[0],a3,a0[1:])&nbs***bsp;add(a5,a3+a0[0],a0[1:])

while 1:
    try:
        n = int(input())
        m = list(map(int,input().split()))
        a = [[],[],[]]
        for i in m:
            if i%5 == 0:
                a[0].append(i)
            elif i%3 == 0:
                a[1].append(i)
            else:
                a[2].append(i)
        a5,a3 = 0,0
        for i in a[0]:
            a5 += i
        for i in a[1]:
            a3 += i
        print(str(add(a5,a3,a[2])).lower())

    except:
        break

n, nums = input(), list(map(int, input().split()))
res = [0]
for i in nums[:]:
    if i % 5 == 0:
        res[0] += i
        nums.remove(i)
    elif i % 3 == 0:
        res[0] -= i
        nums.remove(i)

for i in nums:
    new_res = []
    for e in res:
        new_res.extend([e + i, e - i])
    res = list(set(new_res))
print('true' if 0 in res else 'false')

n = int(input())
s = list(map(int, input().split()))

# a存5的倍数 包括15
a = list()
# b存3的倍数 不包括15 <- 用 elif 实现
b = list()
# c存其他的数
c = list()

for i in range(len(s)):
    if s[i] % 5 == 0:
        a.append(s[i])
    elif s[i] % 3 == 0:
        b.append(s[i])
    else:
        c.append(s[i])
# sum求和
a = sum(a)
b = sum(b)

# 递归一下 
# 虽说条件写的是 k == len(c) 
# 但是其实k=0的时候,第一轮儿计算的时候 abc的值并没有改变
# 第二轮儿的时候 dfs(a+c[k],b,c,k+1)&nbs***bsp; dfs(a,b+c[k],c,k+1) 才改变
# 所以停止条件是 k == len(c) 正好把所有的c里面的值都计算了 a == b时 return True
def dfs(a,b,c,k):
    if k == len(c) and a == b:
        return True 
    if k == len(c) and a != b:
        return False
    res = dfs(a+c[k],b,c,k+1)&nbs***bsp; dfs(a,b+c[k],c,k+1)
    return res  

if dfs(a,b,c,0):
    print('true')
else:
    print('false')

递归非常简单直观

n = input()
nums = list(map(int, list(input().split())))
def group(three, five, others):
    if len(others) == 0:
        if sum(three) == sum(five):
            return True
        else:
            return False
    return group(three+[others[0]], five, others[1:]) or group(three, five+[others[0]], others[1:])

three, five, others = [], [], []
for num in nums:
    if num % 3 == 0:
        three.append(num)
    elif num % 5 == 0:
        five.append(num)
    else:
        others.append(num)
if group(three, five, others):
    print('true')
else:
    print('false')