standard 銀行 put sts double style tchar sample NPU

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 Copy

3
5 0 -5

output Copy

1

input Copy

4
-1 0 1 0

output Copy

2

input Copy

4
1 2 3 -6

output Copy

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家銀行，相鄰兩個銀行可以互相周轉要使得最後每家銀行欠款為0，問最少交換次數。

分析：

　　最多最多需要n-1次，推一下可以發現，1~n中每有一個區間內的和為0，交換次數就減一。所以求一下區間和為0的區間有幾個就行了。

　　然後可以推一下得一個結論就是前綴和中出現次數最多的數就是區間內有幾個區間為0。

　　

///  author:Kissheart  ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e5+10;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
ll n;
map<ll,ll>mp;
int main()
{
    scanf("%lld",&n);
    ll ans=n-1,sum=0;
    for(ll i=1;i<=n;i++)
    {
        ll x;
        scanf("%lld",&x);
        sum+=x;
        mp[sum]++;
        ans=min(ans,n-mp[sum]);
    }
    printf("%lld\n",ans);
    return 0;
}

View Code

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