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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.
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.
Print the minimum number of operations required to change balance in each bank to zero.
3
5 0 -5
1
4
-1 0 1 0
2
4
1 2 3 -6
3
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:
题意：
有n个银行，排成一个圈，相邻两个可以互相转钱，每个银行有一定的钱（可能为负），所有银行的钱加起来为0，现在小A想知道最少通过多少次转钱使得所有银行的钱均为0
分析：
先求前缀和；
如果中间有一段和为0的，假如长度为l，那么次数只要l-1，所以每多一个这样的段总数就可以减少1。
举个栗子：
1 -1 1 2 -3
1~-1 和为零 转1次
1~-3 和为零 转2次
所以说 每次找到和为零的段（长为L)就要转L-1次
答案=n-和为零的段数（互不重叠）
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <map> 6 using namespace std; 7 8 typedef long long LL; 9 map<LL,int> mp; 10 11 int main() 12 { 13 int n; 14 while(cin>>n) 15 { 16 LL a,sum=0; 17 int ans=0; 18 for(int i=1;i<=n;i++) 19 { 20 cin>>a; 21 sum+=a; 22 mp[sum]++; 23 ans=max(ans,mp[sum]); 24 } 25 printf("%d\n",n-ans); 26 } 27 return 0; 28 }
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原文地址：http://www.cnblogs.com/cnblogsLSY/p/5789447.html
