codeforces 675 C ——Money Transfers—————

本文介绍了一道算法题，目标是最少操作次数使各银行账户余额归零。文章提供了问题背景、解决思路及代码实现，并解释了如何通过计算区间和为0的数量来确定最少操作次数。

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

input

4
-1 0 1 0

output

input

4
1 2 3 -6

output

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：给你n个银行中的存款（负值表示借贷），是成环的，1跟n相接，这n个数的和为0。可以从i向i的相邻两侧转移存款，问你最少转移多少次，可以让所有银行的存款都为0。

解题思路：n个数的和为0，假设是由k个和为0的区间组成的。假设这些区间的长度为l_i，那么长度为l_i的区间需要l_i-1次转移。那么总共需要l₁-1+l₂-1+l₃-1...+l_k-1次转移，因为l₁+l₂+l₃...l_k=n，所以总共需要n-k次转移，这个k就是区间和为0的区间个数。我们通过求前缀和sum来统计区间和为0的个数。当sum[i] = sum[j]时，我们知道a[i+1]+...a[j]为0，这个区间和为0。所以我们记录sum=X出现的次数，n-time[X]来更新答案。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int mod = 1e9+7;
const int maxn = 1e4+200;
const int INF = 0x3f3f3f3f;
map<LL, int>time;
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        time.clear();
        LL sum = 0;
        int ans = n-1;
        int val;
        for(int i = 1; i <= n; i++){
            scanf("%d",&val);
            sum += val;
            time[sum]++;
            ans = min(ans,n-time[sum]);
        }
        printf("%d\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/chengsheng/p/5535316.html