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.

講道理這道題現在還是不知道正解是啥。。。

看到了一份能看懂的程式碼，先貼上來。再說我自己的想法。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 500 +10;
const long long INF=0x7fffffffffffffff;
long long ans=INF;
long long a[MAXN];
int n;

void solve(long long x){
    long long s=0;
    if(x==0)
        return;
    for(int i=0;i<n;i++){
        long long p=a[i]/x;
        long long q=a[i]%x;
        if(p==0)
            return;
        if(p>=q){
            s+=(a[i]+x)/(x+1);
        }else
            return;
    }
    ans=min(ans,s);
}



int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%I64d",a+i);
    }
    sort(a,a+n);
    for(int i=1;i*i<=a[0];i++){
        solve(a[0]/i);
        solve(a[0]/i+1);
        solve(i);
        solve(a[0]/i-1);
    }
    printf("%I64d\n",ans);
}

其實就是暴力，因為每一份的個數肯定不能超過最小的數，所以針對最小的數列舉一下。不過在這個程式碼裡面那個a[i]+x/(x+1)不太懂

我是用不定方程做的。

對於ax+by=n,x=x'-bt,y=y'+at.x',y'為一組特解。

對於p=a[I]/x,q=a[I]%x.

只要p>=q，就能把q分成q個1加到其他集合裡面。(p-q)*x+q*(x+1)=a[i],總個數就是p+t,求出t的範圍即可。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 500 +10;
const long long INF=0x7fffffffffffffff;
long long ans=INF;
long long a[MAXN];
int n;

void solve(long long x){
    long long s=0;
    if(x==0)
        return;
    for(int i=0;i<n;i++){
        long long p=a[i]/x;
        long long q=a[i]%x;
        if(p<q){
            return;
        }
        long long X=(a[i]-q)/x;
        long long Y=(p-q)/(x+1);
        long long temp=min(X,Y);
        s+=p-temp;
    }
    ans=min(ans,s);
}


int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%I64d",a+i);
    }
    sort(a,a+n);
    for(int i=1;i*i<=a[0];i++){
        solve(a[0]/i);
        solve(a[0]/i+1);
        solve(i);
        solve(a[0]/i-1);
    }
    printf("%I64d\n",ans);
}

然而居然被卡常樂。。

氣哭