Only divisible Given an array A of N integers, and an integer x. For each element in A you can do the following operatio
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Only divisible Given an array A of N integers, and an integer x. For each element in A you can do the following operatio
Only divisible Given an array A of N integers, and an integer x. For each element in A you can do the following operation at most once : add +1 or -1 to A. (1<i< N). Let S be the summation of elements of some subsequence in A after performing some operations. Find the maximum integer S such that S is divisible by x, Le (S % x = 0). Example Consider N = 3, X = 2, Al] =[1, 2, 3]. • Increase A[1] by 1. we get- A=[2, 2, 3]. • increase A[2] by 1. we get- A=[2, 3, 3]. The Sum of the above array is 8. and it is also divisible by X-2. so, this is the maximum possible value of S. Function description Complete the solve function provided in the editor. This function takes the following 3 parameters and returns the maximum possible S.:
comprene une surve toncuon provided in the comfor. This following 3 parameters and returns the maximum possible S.: • N: Represents the size of the Array. • X: Represents the number. • A: Represents the array N of size N Input format Note: This is the input format that you must use to provide custom input (available above the Compile and Test button). . In the first line, you will be given two integers N and X. • for each of the following N lines, the ith line contains one integer A. Output format • one integer S. Constraints 1<N<100 1≤ A ≤ 109 1<7<105 Code snippets (also called starter code/boilerplate code)
1≤A ≤ 109 1≤#≤ 105 Code snippets (also called starter code/boilerplate code) This question has code snippets for C, CPP, Java, and Python. Sample input D 54 4 8 6 2 1 Sample output 24 Explanation the maximum sum that could be achieved by adding 1 to A[1], A[2], A[3] and A[4], and the answer is the sum of these 4 values. L
espace std; long long Solve (int n, int x, vector<int> a) { } int main() { ios::sync_with_stdio(0); cin.tie(0); int N; cin >> N int X; cin >> X; vector<int> A(N); for(int i_A = 0; i_A < N; i_A++) { cin >> A[i_A]; } long long out_;