P1031 [noip2002 improvement group] average Solitaire

P1031 [NOIP2002 Improvement group ] Divide the cards equally

# [NOIP2002 Improvement group ] Divide the cards equally

## Title Description

Yes $N$ Pile up cards , The numbers are $1,2,…,N$. There are several sheets on each pile , But the total number of cards must be $N$ Multiple . You can take several cards from any pile , And then move .

The transfer rule is ： At No $1$ Cards from the pile , Can only be moved to the number of $2$ On the pile of ; At No $N$ A deck of cards , Can only be moved to the number of $N-1$ On the pile of ; Other cards from the pile , It can be moved to the adjacent left or right pile .

Now we need to find a way to move , Make the same number of cards on each stack with the least number of moves .

for example $N=4$ when ,$4$ The number of cards in the pile is $9,8,17,6$.

Move $3$ Once attainable goal ：

- Take from the third pile $4$ Put cards in the fourth pile , At this time, the number of cards in each stack is $9,8,13,10$.
- Take from the third pile $3$ Put cards in the second pile , At this time, the number of cards in each stack is $9,11,10,10$.
- From the second stack $1$ Put cards in the first pile , At this time, the number of cards in each stack is  $10,10,10,10$.

## Input format

The first line is an integer $N$, Indicates the number of stacks of cards .  
The second line is $N$ It's an integer $A_1,A_2,\cdots,A_N$, Indicates the number of cards at the beginning of each stack .

## Output format

All in one line , That is, the minimum number of moves when all heaps are equal .

## Examples #1

### The sample input #1

```
4
9 8 17 6
```

### Sample output #1

```
3
```

## Tips

about $100\%$ The data of ,$1  \le  N  \le  100$,$1 \le  A_i  \le 10000$.

**【 Title source 】**

NOIP 2002 Improve the first question of the group

【AC Code 】

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<string>
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int N=110;
const double eps=1e-5;
int read()
{
	char ch=getchar();
	int x=0,f=1;
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-48,ch=getchar();}
	return x*f;
}
int n,a[N],avr,ans;
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		a[i]=read();
		avr+=a[i];
	}
	avr/=n;
	for(int i=1;i<=n;i++)
		if(a[i]-avr)
		{
			a[i+1]+=a[i]-avr;
			ans++;
		}
	printf("%d",ans);
	return 0;
}