Codeforces Round 799 (Div. 4) |
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Finished |
Given an array $$$a$$$ of positive integers with length $$$n$$$, determine if there exist three distinct indices $$$i$$$, $$$j$$$, $$$k$$$ such that $$$a_i + a_j + a_k$$$ ends in the digit $$$3$$$.
The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.
The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the length of the array.
The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array.
The sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.
Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. Output "YES" if there exist three distinct indices $$$i$$$, $$$j$$$, $$$k$$$ satisfying the constraints in the statement, and "NO" otherwise.
You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
6420 22 19 8441 11 1 202241100 1100 1100 1111512 34 56 78 9041 9 8 4616 38 94 25 18 99
YES YES NO NO YES YES
In the first test case, you can select $$$i=1$$$, $$$j=4$$$, $$$k=3$$$. Then $$$a_1 + a_4 + a_3 = 20 + 84 + 19 = 123$$$, which ends in the digit $$$3$$$.
In the second test case, you can select $$$i=1$$$, $$$j=2$$$, $$$k=3$$$. Then $$$a_1 + a_2 + a_3 = 1 + 11 + 1 = 13$$$, which ends in the digit $$$3$$$.
In the third test case, it can be proven that no such $$$i$$$, $$$j$$$, $$$k$$$ exist. Note that $$$i=4$$$, $$$j=4$$$, $$$k=4$$$ is not a valid solution, since although $$$a_4 + a_4 + a_4 = 1111 + 1111 + 1111 = 3333$$$, which ends in the digit $$$3$$$, the indices need to be distinct.
In the fourth test case, it can be proven that no such $$$i$$$, $$$j$$$, $$$k$$$ exist.
In the fifth test case, you can select $$$i=4$$$, $$$j=3$$$, $$$k=1$$$. Then $$$a_4 + a_3 + a_1 = 4 + 8 + 1 = 13$$$, which ends in the digit $$$3$$$.
In the sixth test case, you can select $$$i=1$$$, $$$j=2$$$, $$$k=6$$$. Then $$$a_1 + a_2 + a_6 = 16 + 38 + 99 = 153$$$, which ends in the digit $$$3$$$.
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