A Simple Stone Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1526    Accepted Submission(s): 346

Problem Description

After he has learned how to play Nim game, Bob begins to try another stone game which seems much easier.

The game goes like this: one player starts the game with N piles of stones. There is a_i stones on the ith
pile. On one turn, the player can move exactly one stone from one pile
to another pile. After one turn, if there exits a number x(x > 1) such that for each pile b_i is the multiple of x where b_i
is the number of stone of the this pile now), the game will stop. Now
you need to help Bob to calculate the minimum turns he need to stop this
boring game. You can regard that 0 is the multiple of any positive number.

Input

The first line is the number of test cases. For each test case, the first line contains one positive number N(1 \leq N \leq 100000), indicating the number of piles of stones.

The second line contains N positive number, the ith number a_i (1 \leq a_i \leq 100000) indicating the number of stones of the ith pile.

The sum of N of all test cases is not exceed 5 * 10^5.

Output

For each test case, output a integer donating the answer as described above. If there exist a satisfied number x initially, you just need to output 0. It‘s guaranteed that there exists at least one solution.

Sample Input

2
5
1 2 3 4 5
2
5 7

Sample Output

2
1

/*************************************************************************
    > File Name: H.cpp
    > Author: LyuCheng
    > Created Time: 2017-12-01 20:41
    > Description:
        题意：有n堆石子，每次你可以选择一堆石子拿一个放到另一堆石子里，如果
            所有的所有的石子数不互质，那么游戏结束
        思路：问题的实质就是找到一个x是的b[1]%x+b[2]%x+...b[n]%x=0;也就是
            sum%x=0,那么分解sum的质因子,然后枚举判断
 ************************************************************************/

#include <bits/stdc++.h>

#define MAXN 123456
#define LL long long
#define INF 50000000005 

using namespace std;

bool prime[MAXN];
LL p[MAXN];
LL tol;
int t;
int n;
LL a[MAXN];
LL g;
LL pos;
LL fa[MAXN];
LL sum;
LL mod;
LL res;
vector<LL>v;

inline void pre(){
    for(LL i=2;i<MAXN;i++){
        if(prime[i]==false)
            p[tol++]=i;
        for(LL j=0;j<tol&&i*p[j]<MAXN;j++){
            prime[i*p[j]]=true;
            if(i%p[j]==0)
                break;
        }
    }
}

inline void div(LL x){
    memset(fa,0,sizeof fa);
    pos=0;
    for(LL i=0;i<tol&&p[i]*p[i]<=x;i++){
        if(x%p[i]==0){
            fa[pos++]=p[i];
            while(x%p[i]==0) x/=p[i];
        }
    }
    if(x>1)    fa[pos++]=x;
}

inline void init(){
    g=0;
    sum=0;
    v.clear();
    res=INF;
}

int main(){
    pre();
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(LL i=0;i<n;i++){
            scanf("%lld",&a[i]);
            sum+=a[i];
            g=__gcd(a[i],g);
        }
        if(n==1){
            puts("0");
            continue;
        }
        div(sum);
        for(LL i=0;i<pos;i++){
            v.clear();
            mod=fa[i];
            LL cnt=0;
            for(LL i=0;i<n;i++){
                if(a[i]%mod!=0){
                    v.push_back(a[i]%mod);
                    cnt+=a[i]%mod;
                }
            }
            sort(v.begin(),v.end());
            LL tol=cnt/mod;
            LL s=0;
            for(LL i=(LL)v.size()-1;i>=0;i--){
                tol--;
                s+=(LL)(mod-v[i]);
                if(tol<=0) break;
            }
            res=min(res,s);
        }
        printf("%lld\n",res);
    }
    return 0;
}