题解 | 九倍平方数

def solve(num):
    sun = sum(num)
    if sun % 9 == 0:
        return "YES"
    n2 = num.count(2)#统计输入数字中，2，3的个数，方便进行变换操作
    n3 = num.count(3)
    if n2 == 0 and n3 == 0:#分四种情况进行讨论
        return "NO"
    elif n2 != 0 and n3 == 0:
        for i in range(n2):
            if (2 + sun + i * 2) % 9 == 0:
                return "YES"
        return "NO"
    elif n2 == 0 and n3 != 0:
        for i in range(n3):
            if (6 + sun + i * 6) % 9 == 0:
                return "YES"
        return "NO"
    else:
        for i in range(n2 + 1):
            for j in range(n3 + 1):
                if (sun + i * 2 + j * 6) % 9 == 0:
                    return "YES"
        return "NO"

#整数能被9整除，即各项数字的和能被9整除
#0，1，2，3的平方小于10。0，1的平方不改变和，2，3的平方使和加4，6
t = int(input())
for _ in range(t):
    num = list(map(int, list(input())))
    print(solve(num))