Codeforces Round 690 (Div. 3)

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1462 Codeforces Round 690 (Div. 3) FINISHED False 7200 123780299 Dec. 15, 2020, 2:35 p.m.

Problems

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( 9840 ) E2 Close Tuples (hard version) PROGRAMMING binary search combinatorics implementation math sortings two pointers

B'This is the hard version of this problem. The only difference between the easy and hard versions is the constraints on k and m . In this version of the problem, you need to output the answer by modulo 10^9+7 . You are given a sequence a of length n consisting of integers from 1 to n . The sequence may contain duplicates (i.e. some elements can be equal). Find the number of tuples of m elements such that the maximum number in the tuple differs from the minimum by no more than k . Formally, you need to find the number of tuples of m indices i_1 < i_2 < ldots < i_m , such that max(a_{i_1}, a_{i_2}, ldots, a_{i_m}) - min(a_{i_1}, a_{i_2}, ldots, a_{i_m}) <= k. For example, if n=4 , m=3 , k=2 , a=[1,2,4,3] , then there are two such triples ( i=1, j=2, z=4 and i=2, j=3, z=4 ). If n=4 , m=2 , k=1 , a=[1,1,1,1] , then all six possible pairs are suitable. As the result can be very large, you should print the value modulo 10^9 + 7 (the remainder when divided by 10^9 + 7 ). The first line contains a single integer t ( 1 <= t <= 2 cdot 10^5 ) -- the number of test cases. Then t test cases follow. The first line of each test case contains three integers n , m , k ( 1 <= n <= 2 cdot 10^5 , 1 <= m <= 100 , 1 <= k <= n ) -- the length of the sequence a , number of elements in the tuples and the maximum difference of elements in the tuple. The next line contains n integers a_1, a_2, ldots, a_n ( 1 <= a_i <= n ) -- the sequence a . It is guaranteed that the sum of n for all test cases does not exceed 2 cdot 10^5 . Output t answers to the given test cases. Each answer is the required number of tuples of m elements modulo 10^9 + 7 , such that the maximum value in the tuple differs from the minimum by no more than k .'...

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Codeforces Round #690 (Div. 3) Editorial

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