SRM 398 DIV2 [250]

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

class MinDifference...{
public:
    int closestElements(int A0,int X,int Y,int M,int n)
    ...{
        vector<int> ivec(n);
        vector<int>::iterator iter=ivec.begin();
        *iter++=A0;
        for (;iter!=ivec.end();++iter)
            *iter=(*(iter-1)*X+Y)%M;
        sort(ivec.begin(),iter);
        iter=ivec.begin()+1;
        int min=*iter-*(iter-1);
        for (;iter!=ivec.end();++iter)
            if (*iter-*(iter-1)<min) min=*iter-*(iter-1);
        return min;
    }
};

Problem Statement
	You are given numbers A0, X, Y, M and n. Generate a list A of length n according to the following recurrence relation: A[0] = A0 A[i] = (A[i - 1] * X + Y) MOD M, for 0 < i < n Return the minimal absolute difference between any two elements of A.
Definition
	Class: MinDifference Method: closestElements Parameters: int, int, int, int, int Returns: int Method signature: int closestElements(int A0, int X, int Y, int M, int n) (be sure your method is public)
Constraints
-	A0, X, Y, M will each be between 1 and 10000, inclusive.
-	n will be between 2 and 10000, inclusive.
Examples
0)
	3 7 1 101 5 Returns: 6 The elements of the list are {3, 22, 54, 76, 28}. The minimal difference is between elements 22 and 28.
1)
	3 9 8 32 8 Returns: 0 All elements are the same.
2)
	67 13 17 4003 23 Returns: 14
3)
	1 1221 3553 9889 11 Returns: 275
4)
	1 1 1 2 10000 Returns: 0
5)
	1567 5003 9661 8929 43 Returns: 14

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