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Discussion of Problem 1032. Find a Multiple

Explanation why continuous segment is always there whose sum is divisible by n
Posted by Vineet Jain 7 Dec 2020 23:59
Let suppose there is prefix sum array with mod n is pre1,pre2,......,pren. Since there can be only n-1 numbers present except zero. If zero is one of elements in prefix sum array then answer is already 0 to i. Else If zero is not present then some number should repeat because prefix array is of size n but only 1,2.... n-1 numbers are present. So there should be particular i,j => prei==prej. Therefore from i+1 to j segment will be divisible by n.

Edited by author 08.12.2020 00:01