P16746 [GKS 2019 #H] Elevanagram

It is a well known fact that a number is divisible by $11$ if and only if the alternating sum of its digits is equal to $0$ modulo $11$. For example, $8174958$ is a multiple of $11$, since $8 - 1 + 7 - 4 + 9 - 5 + 8 = 22$. Given a number that consists of digits from $1$-$9$, can you rearrange the digits to create a number that is divisible by $11$? Since the number might be quite large, you are given integers $A_1, A_2, ..., A_9$. There are $A_i$ digits i in the number, for all i.

The first line of the input gives the number of test cases, $T$. $T$ lines follow. Each line contains the nine integers $A_1, A_2, ..., A_9$.

For each test case, output one line containing Case #x: y, where x is the test case number (starting from $1$) and y is **YES** if the digits can be rearranged to create a multiple of $11$, and **NO** otherwise.

- In Sample Case #1, the digits are $336$, which can be rearranged to $363$. This is a multiple of $11$ since $3 - 6 + 3 = 0$. - In Sample Case #2, the digits are $999999999999$, which is already a multiple of $11$, since $9 - 9 + 9 - 9 + ... - 9 = 0$. - In Sample Case #3, the digits are $5578$, which cannot be rearranged to form a multiple of $11$. - In Sample Case #4, the digits are $111234$, which can be rearranged to $142131$. This is a multiple of $11$ since $1 - 4 + 2 - 1 + 3 - 1 = 0$. - In Sample Case #5, the digits are $11177799$, which can be rearranged to $19191777$. This is a multiple of $11$ since $1 - 9 + 1 - 9 + 1 - 7 + 7 - 7 = -22$ (which is $0$ modulo $11$). - In Sample Case #6, the only digit is $8$, which cannot be rearranged to form a multiple of $11$. ### Limits $1 \le T \le 100$. $1 \le A_1 + A_2 + ... + A_9$. **Test set 1 (Visible)** $0 \le A_i \le 20$, for all i. **Test set 2 (Hidden)** $0 \le A_i \le 10^9$, for all i.