CF1692F.3SUM

普及/提高-

通过率:0%

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题目描述

Given an array aa of positive integers with length nn , determine if there exist three distinct indices ii , jj , kk such that ai+aj+aka_i + a_j + a_k ends in the digit 33 .

输入格式

The first line contains an integer tt ( 1t10001 \leq t \leq 1000 ) — the number of test cases.

The first line of each test case contains an integer nn ( 3n21053 \leq n \leq 2 \cdot 10^5 ) — the length of the array.

The second line of each test case contains nn integers a1,a2,,ana_1, a_2, \dots, a_n ( 1ai1091 \leq a_i \leq 10^9 ) — the elements of the array.

The sum of nn across all test cases does not exceed 21052 \cdot 10^5 .

输出格式

Output tt lines, each of which contains the answer to the corresponding test case. Output "YES" if there exist three distinct indices ii , jj , kk 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).

输入输出样例

  • 输入#1

    6
    4
    20 22 19 84
    4
    1 11 1 2022
    4
    1100 1100 1100 1111
    5
    12 34 56 78 90
    4
    1 9 8 4
    6
    16 38 94 25 18 99

    输出#1

    YES
    YES
    NO
    NO
    YES
    YES

说明/提示

In the first test case, you can select i=1i=1 , j=4j=4 , k=3k=3 . Then a1+a4+a3=20+84+19=123a_1 + a_4 + a_3 = 20 + 84 + 19 = 123 , which ends in the digit 33 .

In the second test case, you can select i=1i=1 , j=2j=2 , k=3k=3 . Then a1+a2+a3=1+11+1=13a_1 + a_2 + a_3 = 1 + 11 + 1 = 13 , which ends in the digit 33 .

In the third test case, it can be proven that no such ii , jj , kk exist. Note that i=4i=4 , j=4j=4 , k=4k=4 is not a valid solution, since although a4+a4+a4=1111+1111+1111=3333a_4 + a_4 + a_4 = 1111 + 1111 + 1111 = 3333 , which ends in the digit 33 , the indices need to be distinct.

In the fourth test case, it can be proven that no such ii , jj , kk exist.

In the fifth test case, you can select i=4i=4 , j=3j=3 , k=1k=1 . Then a4+a3+a1=4+8+1=13a_4 + a_3 + a_1 = 4 + 8 + 1 = 13 , which ends in the digit 33 .

In the sixth test case, you can select i=1i=1 , j=2j=2 , k=6k=6 . Then a1+a2+a6=16+38+99=153a_1 + a_2 + a_6 = 16 + 38 + 99 = 153 , which ends in the digit 33 .

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