Codeforces Round #353 (Div. 2) C. Money Transfers (map)

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Codeforces Round #353 (Div. 2) C. Money Transfers (map)

C. Money Transfers

time limit per test

memory limit per test

input

output

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

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 (9 ≤ 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

35 0 -5

output

1

input

4-1 0 1 0

output

2

input

41 2 3 -6

output

3

Note

In the first sample, Vasya may transfer 5

In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1

In the third sample, the following sequence provides the optimal answer:

transfer 1transfer 3transfer 6

题解:有n个负担着不同债务的银行围成环,每次每个银行只能向自己左边或者右边的银行转移资金。所有银行的债务和为0,问你最少转移多少次能让所有银行债务都为0。

map一下就好了。其实就是前缀和出现次数最多的就是区间和为0的区间宽度。

AC代码:

#pragma comment(linker, "/STACK:102400000,102400000")//#include#include#include#include#include#include#include#include#include#include#include#include#include using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;typedef vector vi;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)#define in freopen("in.txt","r",stdin) #define rep(i,j,k) for (int i = j; i <= k; i++) #define per(i,j,k) for (int i = j; i >= k; i--) #define lson x << 1, l, mid #define rson x << 1 | 1, mid + 1, r const int lowbit(int x) { return x&-x; } const double eps = 1e-8; const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e6 + 3; const ll mod = (1LL<<32);const int N =1e6+6; const int M=100010;//const int maxn=1001; template inline void getmax(T1 &a, T2 b) {if (b>a)a = b;} template inline void getmin(T1 &a, T2 b) {if (bmp;int main(){ ll x,y; cin>>n; x=0; for(ll i=0;i>y; maxn=max(maxn,++mp[x+=y]); } cout<

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