codeforces 675C Money Transfers

codeforces 675C Money Transfers

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 a_i ( - 10⁹ ≤ a_i ≤ 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 a_i is equal to 0.

Output

Print the minimum number of operations required to change balance in each bank to zero.

Sample Input

3
5 0 -5

1

4
-1 0 1 0

2

4
1 2 3 -6

3

题意：N个银行构成一个环，已知每个银行的的金钱数，求最少移动多少步使得所有的钱变为0

思路：假定最坏的情况，就是都移动到最后的银行里面钱数菜户变为0那就需要N-1次 ，中途每遇到一次0说明次数减少一次，所以只需要统计一次0的个数即可。

AC代码：

<span style="font-size:14px;">#include <cstring>
#include <queue>
#include <vector>
#include <map>using namespace std;int n;
typedef long long ll;
map<ll,int>q;int main(){cin >> n;ll sum = 0;int tol = 0;int g = n;while(n--){long long a;cin >> a;sum += a;q[sum]++;tol = max(q[sum],tol);}printf("%dn",g-tol);return  0;
}</span>