CF1789E Serval and Music Game

Serval loves playing music games. He meets a problem when playing music games, and he leaves it for you to solve. You are given $ n $ positive integers $ s_1 < s_2 < \ldots < s_n $ . $ f(x) $ is defined as the number of $ i $ ( $ 1\leq i\leq n $ ) that exist non-negative integers $ p_i, q_i $ such that: $$ s_i=p_i\left\lfloor{s_n\over x}\right\rfloor + q_i\left\lceil{s_n\over x}\right\rceil $$ Find out $ \sum_{x=1}^{s_n} x\cdot f(x)$ modulo $ 998\,244\,353 $ . As a reminder, $ \lfloor x\rfloor $ denotes the maximal integer that is no greater than $ x $ , and $ \lceil x\rceil $ denotes the minimal integer that is no less than $x$.

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1\leq t\leq 10^4 $ ). The description of the test cases follows. The first line of each test cases contains a single integer $ n $ ( $ 1\leq n\leq 10^6 $ ). The second line of each test case contains $ n $ positive integers $ s_1,s_2,\ldots,s_n $ ( $ 1\leq s_1 < s_2 < \ldots < s_n \leq 10^7 $ ). It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 10^6 $ , and the sum of $ s_n $ does not exceed $ 10^7 $ .

For each test case, print a single integer in a single line — the sum of $ x\cdot f(x) $ over all possible $ x $ modulo $ 998\,244\,353 $ .

For the first test case, $ s_n=4 $ , $ f(x) $ are calculated as followings: - $ f(1)=1 $ - $ \left\lfloor s_n\over 1\right\rfloor=4 $ , $ \left\lceil s_n\over 1\right\rceil=4 $ . - It can be shown that such $ p_1,p_2 $ and $ q_1,q_2 $ that satisfy the conditions don't exist. - Let $ p_3=1 $ and $ q_3=0 $ , then $ s_3 = p_3\cdot\left\lfloor s_n\over 1\right\rfloor + q_3\cdot\left\lceil s_n\over 1\right\rceil = 1\cdot 4 + 0\cdot 4 = 4 $ . - $ f(2)=2 $ - $ \left\lfloor s_n\over 2\right\rfloor=2 $ , $ \left\lceil s_n\over 2\right\rceil=2 $ . - It can be shown that such $ p_1 $ and $ q_1 $ that satisfy the conditions don't exist. - Let $ p_2=1 $ and $ q_2=0 $ , then $ s_2 = p_2\cdot\left\lfloor s_n\over 2\right\rfloor + q_2\cdot\left\lceil s_n\over 2\right\rceil = 1\cdot 2 + 0\cdot 2 = 2 $ . - Let $ p_3=0 $ and $ q_3=2 $ , then $ s_3 = p_3\cdot\left\lfloor s_n\over 2\right\rfloor + q_3\cdot\left\lceil s_n\over 2\right\rceil = 0\cdot 2 + 2\cdot 2 = 4 $ . - $ f(3)=3 $ - $ \left\lfloor s_n\over 3\right\rfloor=1 $ , $ \left\lceil s_n\over 3\right\rceil=2 $ . - Let $ p_1=1 $ and $ q_1=0 $ , then $ s_1 = p_1\cdot\left\lfloor s_n\over 3\right\rfloor + q_1\cdot\left\lceil s_n\over 3\right\rceil = 1\cdot 1 + 0\cdot 2 = 1 $ . - Let $ p_2=0 $ and $ q_2=1 $ , then $ s_2 = p_2\cdot\left\lfloor s_n\over 3\right\rfloor + q_2\cdot\left\lceil s_n\over 3\right\rceil = 0\cdot 1 + 1\cdot 2 = 2 $ . - Let $ p_3=0 $ and $ q_3=2 $ , then $ s_3 = p_3\cdot\left\lfloor s_n\over 3\right\rfloor + q_3\cdot\left\lceil s_n\over 3\right\rceil = 0\cdot 1 + 2\cdot 2 = 4 $ . - $ f(4)=3 $ - $ \left\lfloor s_n\over 4\right\rfloor=1 $ , $ \left\lceil s_n\over 4\right\rceil=1 $ . - Let $ p_1=1 $ and $ q_1=0 $ , then $ s_1 = p_1\cdot\left\lfloor s_n\over 4\right\rfloor + q_1\cdot\left\lceil s_n\over 4\right\rceil = 1\cdot 1 + 0\cdot 1 = 1 $ . - Let $ p_2=1 $ and $ q_2=1 $ , then $ s_2 = p_2\cdot\left\lfloor s_n\over 4\right\rfloor + q_2\cdot\left\lceil s_n\over 4\right\rceil = 1\cdot 1 + 1\cdot 1 = 2 $ . - Let $ p_3=2 $ and $ q_3=2 $ , then $ s_3 = p_3\cdot\left\lfloor s_n\over 4\right\rfloor + q_3\cdot\left\lceil s_n\over 4\right\rceil = 2\cdot 1 + 2\cdot 1 = 4 $ . Therefore, the answer is $ \sum_{x=1}^4 x\cdot f(x) = 1\cdot 1 + 2\cdot 2 + 3\cdot 3 + 4\cdot 3 = 26 $ . For the second test case: - $ f(1)=f(2)=f(3)=1 $ - $ f(4)=3 $ - $ f(5)=f(6)=f(7)=f(8)=f(9)=4 $ For example, when $ x=3 $ we have $ f(3)=1 $ because there exist $ p_4 $ and $ q_4 $ : $$ 9 = 1 \cdot\left\lfloor{9\over 3}\right\rfloor + 2 \cdot\left\lceil{9\over 3}\right\rceil $$ It can be shown that it is impossible to find $ p_1,p_2,p_3 $ and $ q_1,q_2,q_3 $ that satisfy the conditions. When $ x=5 $ we have $ f(5)=4 $ because there exist $ p_i $ and $ q_i $ as followings: $$ 1 = 1 \cdot\left\lfloor{9\over 5}\right\rfloor + 0 \cdot\left\lceil{9\over 5}\right\rceil $$ $$ 2 = 0 \cdot\left\lfloor{9\over 5}\right\rfloor + 1 \cdot\left\lceil{9\over 5}\right\rceil $$ $$ 7 = 3 \cdot\left\lfloor{9\over 5}\right\rfloor + 2 \cdot\left\lceil{9\over 5}\right\rceil $$ $$ 9 = 3 \cdot\left\lfloor{9\over 5}\right\rfloor + 3 \cdot\left\lceil{9\over 5}\right\rceil $$ Therefore, the answer is $ \sum_{x=1}^9 x\cdot f(x) = 158$.