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

C. Money Transfers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 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: 

1. transfer 1 from the first bank to the second bank; 
2. transfer 3 from the second bank to the third; 
3. transfer 6 from the third bank to the fourth. 

題意：

n個負擔着不同債務的銀行圍成環，每次每個銀行只能向自己左邊或者右邊的銀行轉移資金。所有銀行的債務和爲0，問你最少轉移多少次能讓所有銀行債務都爲0。

分析：

n個銀行，最多n-1次即可將所有銀行債務變爲0，即按順序從第一個開始向後一個銀行轉移債務即可，到第n個時所有債務必爲0。

在此轉移過程中可以發現任何一個爲0的區間都可以減少1次轉移，即該區間和爲0時，右邊界不需要向後一個銀行繼續轉移債務。

數出存在多少個0區間即可得到答案爲n-0區間個數。

假設區間[i,j]和爲0，那麼[1,i-1]的區間和與[1,j]和相同。

通過map存下區間和大小出現的次數，即可得到哪些區間爲0。

但從1-n遞推時是否沒考慮到環？由於計算[1,x]（x從1-n）的區間和並存入map中，因此存在環的情況並不影響和的出現。

思路來自：http://m.blog.csdn.net/article/details?id=51438865

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <fstream>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define INF 0x3f3f3f3f
const int N=100005;
const int mod=1e9+7;

int a;
map<long long, int> findsum;

int main() {
    int n;
    long long sum=0;
    int ans=0;
    cin>>n;
    for (int i=0; i<n; i++) {
        scanf("%d",&a);
        sum+=a;
        ans=max(ans, ++findsum[sum]);
    }
    cout<<n-ans<<endl;
    return 0;
}