题解 | #数组分组#

# 深度优先递归
# 题目要求将数组分为两组，5和3的倍数不能在同一组内，非5或3的倍数随意放在任意一组，要求分配之后两组和相等
# 根据题意将数组中的5和3分别放到两组中，然后递归穷举所有非5或3的倍数的分组情况
def dfs(three, five, other):
    if not other:
        if sum(three) == sum(five):
            return True
        else:
            return False
    if dfs(three+other[:1], five, other[1:]):
        return True
    if dfs(three, five+other[:1], other[1:]):
        return True

while True:
    try:
        n, nums = int(input()), list(map(int, input().split()))
        three, five, other = [], [], []
        for num in nums:
            if num % 3 == 0:
                three.append(num)
            elif num % 5 == 0:
                five.append(num)
            else:
                other.append(num)
        if dfs(three, five, other):
            print ('true')
        else:
            print ('false')
    except:
        break