MKEQUAL

Logic : 

            we can make N - 1 equal easily by choosing a single element for decreasing.

             so according to first statement answer will be atleast N - 1 now if see this test case

             

             N = 4   Arr = {1, 9, 1, 9}
             answer is 4 because we can make 5 by choosing 1, 9 as pair and perform operation.

             

             that mean answer can be N also , now question is when it can be happen and why ?

             

 answer of above question is :

            

              If sum of numbers is divisible by n then all element changed to sum_element/n by 

              choosing two  index one with value greater than sum_element/n and other with

              smaller value at a time.

Below you can see code also

#include <iostream>

#include <sstream>

#include <cstdio>

#include <cmath>

#include <cstring>

#include <cctype>

#include <string>

#include <vector>

#include <list>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <algorithm>

#include <functional>

using namespace std; 

#define DEBUG(x) cout << '>' << #x << ':' << x << endl;

#define REP(i,n) for(int i=0;i<(n);i++)

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)

#define FORD(i,a,b) for(int i=(a);i>=(b);i--)

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }

const int INF = 1<<29;

typedef long long ll;

inline int two(int n) { return 1 << n; }

inline int test(int n, int b) { return (n>>b)&1; }

inline void set_bit(int & n, int b) { n |= two(b); }

inline void unset_bit(int & n, int b) { n &= ~two(b); }

inline int last_bit(int n) { return n & (-n); }

inline int ones(int n) { int res = 0; while(n && ++res) n-=n&(-n); return res; }

template<class T> void chmax(T & a, const T & b) { a = max(a, b); }

template<class T> void chmin(T & a, const T & b) { a = min(a, b); }

/////////////////////////////////////////////////////////////////////(

int main()

{ 

int T;

scanf("%d", &T);

while(T--) {

int N;

scanf("%d", &N);

long long sum = 0;

for(int i = 0; i < N; ++i) {

long long data;

scanf("%lld", &data);

sum += data;

}

printf("%d\n", N - (sum%N == 0 ? 0 : 1));

}

    return 0;

}