题干：
There are n banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank n are neighbours if n > 1. No bank is a neighbour of itself.
Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.
There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.
Vasya doesn't like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It's guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of banks.
The second line contains n integers ai ( - 109 ≤ ai ≤ 109), the i-th of them is equal to the initial balance of the account in the i-th bank. It's guaranteed that the sum of all ai is equal to 0.
Output
Print the minimum number of operations required to change balance in each bank to zero.
Examples
Input
3
5 0 -5
Output
1
Input
4
-1 0 1 0
Output
2
Input
4
1 2 3 -6
Output
3
Note
In the first sample, Vasya may transfer 5 from the first bank to the third.
In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1 from the second to the first.
In the third sample, the following sequence provides the optimal answer:
- transfer 1 from the first bank to the second bank;
- transfer 3 from the second bank to the third;
- transfer 6 from the third bank to the fourth.
题目大意：
有n家银行，构成了一个环，每个银行都有余额，有的为正数，有的是负数，但是总的余额是0，定义一次操作：相邻的银行之间可以转任意数目的钱。求把所有的银行的余额都转为0的所需最少操作次数。
解题报告：
刚开始是想的以0为分界线，后来发现好像不太行。参考题解
当有一小段(长度为len)的总和为0时, 我们可以用len-1步来使这一小段使这一小段全部变成0,
考虑将n个银行分成k小段, 由于是环型结构并且总和为0, 不会存在最后凑不成总和为0的一小段的情况.
k小段的长度分别是len1,len2,.....lenk, 所需要的步数分别是len1-1,len2-1,....lenk-1, len1+len2+...+lenk
是等于n的, 然后k个-1, 最后就是n-k, 要使n-k最小, k要尽量大. 就是要把n个数分成尽量多的段数.
考虑前缀和, 统计不同前缀和的个数, 相同的前缀和之间夹的一小段数肯定是总和为0的, 用不同的前缀和
划分这n个数k可能不同, 我们统计这些前缀和并且取最大的.
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
map<ll,ll> mp;
int n;
ll x,ans,sum;
int main()
{cin>>n;for(int i = 1; i<=n; i++) {cin>>x;sum += x;mp[sum]++;ans = max(ans,mp[sum]);}cout << n-ans << endl;return 0 ;
}
