
ZOJ Problem Set  4036
BaoBao has just found a positive integer sequence $a_1, a_2, \dots, a_n$ of length $n$ from his left pocket and another positive integer $b$ from his right pocket. As number 7 is BaoBao's favorite number, he considers a positive integer $x$ lucky if $x$ is divisible by 7. He now wants to select an integer $a_k$ from the sequence such that $(a_k+b)$ is lucky. Please tell him if it is possible. InputThere are multiple test cases. The first line of the input is an integer $T$ (about 100), indicating the number of test cases. For each test case: The first line contains two integers $n$ and $b$ ($1 \le n, b \le 100$), indicating the length of the sequence and the positive integer in BaoBao's right pocket. The second line contains $n$ positive integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), indicating the sequence. OutputFor each test case output one line. If there exists an integer $a_k$ such that $a_k \in \{a_1, a_2, \dots, a_n\}$ and $(a_k + b)$ is lucky, output "Yes" (without quotes), otherwise output "No" (without quotes). Sample Input4 3 7 4 5 6 3 7 4 7 6 5 2 2 5 2 5 2 4 26 100 1 2 4 Sample OutputNo Yes Yes Yes HintFor the first sample test case, as 4 + 7 = 11, 5 + 7 = 12 and 6 + 7 = 13 are all not divisible by 7, the answer is "No". For the second sample test case, BaoBao can select a 7 from the sequence to get 7 + 7 = 14. As 14 is divisible by 7, the answer is "Yes". For the third sample test case, BaoBao can select a 5 from the sequence to get 5 + 2 = 7. As 7 is divisible by 7, the answer is "Yes". For the fourth sample test case, BaoBao can select a 100 from the sequence to get 100 + 26 = 126. As 126 is divisible by 7, the answer is "Yes". Author: WENG, Caizhi Source: The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple 