HDU-6237：A Simple Stone Game（数论）

A Simple Stone Game

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
                                                                                             Total Submission(s): 116    Accepted Submission(s): 25

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 ai 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 bi is the multiple of x where bi 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≤N≤100000), indicating the number of piles of stones.

The second line contains N positive number, the ith number ai(1≤ai≤100000) indicating the number of stones of the ith pile.


The sum of N of all test cases is not exceed 5∗105.

 

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

251 2 3 4 525 7

 

Sample Output

21

思路：可以考虑枚举石头数量总和的质因子x，更新答案。对于一个因子x，分别求出a[i]%x的值，那么最少移动次数就相当于把Σ(a[i]%x)剩下的石头移动到Σ(a[i]%x)/x个石堆中的次数，那么当然是选前(a[i]%x)/x大的石堆，然后把其它剩下的石头移到这个前(a[i]%x)/x大的石堆中，即ans=Σ(a[i]%x)-a[i]%x中前k大的a[i]%x。

#include<bits/stdc++.h>using namespace std;const int MAX=1e5+10;long long a[MAX];int all=0,n;int pr[MAX+100];        //素数表bool isp[MAX+10];void init(){    all = 0;    memset(isp,0,sizeof isp);    isp[1]=1;    for(int i=2;i<=MAX;i++)    {        if(!isp[i])pr[all++] = i;        for(int j=0;j<all;j++)        {            long long t = 1LL*pr[j]*i ;            if(t<=MAX)            {                isp[t] = true;                if(i%pr[j]==0)break;            }            else break;        }    }    return ;}struct cmp{    bool operator() (const int a,const int b) const{return a<b;}};long long slove(int prim){    long long sum=0;    priority_queue<int,vector<int>,cmp>p;    for(int i=1;i<=n;i++)    {        sum+=a[i]%prim;        if(a[i]%prim)p.push(a[i]%prim);    }    int t=sum/prim;    while(t--)    {        sum-=p.top();        p.pop();    }    return sum;}int main(){    init();    int T;cin>>T;    while(T--)    {        scanf("%d",&n);        long long sum=0;        for(int i=1;i<=n;i++)scanf("%lld",&a[i]),sum+=a[i];        sort(a+1,a+n+1);        long long ans=sum-a[n];        for(int i=0;i<all;i++)        {            if(sum%pr[i]==0)ans=min(ans,slove(pr[i]));        }        printf("%lld\n",ans);    }    return 0;}