【CF contest/792/problem/E】

E. Colored Balls

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:

• each ball belongs to exactly one of the sets,
• there are no empty sets,
• there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
• there are no two sets such that the difference between their sizes is greater than 1.

Print the minimum possible number of sets.

Input

The first line contains one integer number n (1 ≤ n ≤ 500).

The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).

Output

Print one integer number — the minimum possible number of sets.

Examples

input

3
4 7 8

output

5

input

2
2 7

output

4

Note

In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.

【分析】

　　先%一下大颓果。。

　　这种题自己想分配方法都要对拍一下验证啊不然很容易错。。

　　假设你分成x和x+1，对于某个A来说，判断是否成立的条件是A mod x <= [A / x] 或者 A mod (x+1)==0嘛。。

　　后面那个直接判断就好了，考虑前面那个代表什么。

　　就是A-[A/x]*x<=[A/x]

　　即A/(x+1)<=[A/x] 即 [A/(x+1)]<[A/x]

　　就是说A/(x+1)和A/x要不一样。

　　我们去取最小的A，不同的A/x 只有2根号n个，枚举然后for一遍判断就好了。

　　然后你求出x最大多少，算ans。【我就是这里错了TAT

　　若ans%(x+1)==0 ，直接分成x+1份。

　　否则先分成x份，A%x放入每堆中，剩下可能还有很多堆x，每（x+1）堆x的可以变成x堆（x+1）的，所以减掉 剩下的/（x+1），求出答案即可。

　　时间（根号A * n）

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define Maxn 510
#define INF 0x7fffffff
#define LL long long

int mymin(int x,int y) {return x<y?x:y;}
int mymax(int x,int y) {return x>y?x:y;}

int a[Maxn],n;

bool check(int x)
{
    for(int i=1;i<=n;i++)
    {
        bool p=0;
        if(a[i]%x==0||a[i]%(x+1)==0||a[i]/x!=a[i]/(x+1)) p=1;
        if(!p) return 0;
    }
    return 1;
}

int main()
{
    // int T;
    // scanf("%d",&T);
    // while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        int mn=INF,ans=0;
        for(int i=1;i<=n;i++) mn=mymin(mn,a[i]);
        for(int i=mn;i>=1;)
        {
            if(i<ans) break;
            int x=mn/i,r=mn/(x+1);
            if(check(i)) ans=mymax(ans,i);
            i=r;
        }
        for(int i=1;i<=mn;)
        {
            if(mn/i<ans) break;
            int x=mn/i,r=mn/x;
            if(check(x)) ans=mymax(ans,x);
            else if(check(x-1)) ans=mymax(ans,x-1);
            i=r+1;
        }
        LL sm=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]%(ans+1)==0) sm+=a[i]/(ans+1);
            else
            {
                int xx=a[i]/ans,yy=a[i]%ans;
                // xx-=yy;
                sm+=xx-(xx-yy)/(ans+1);
            }
            // sm+=(a[i]/(ans+1))+(a[i]%(ans+1)!=0);
        }
        printf("%I64d
",sm);
    }
    return 0;
}

2017-04-19 10:02:16