codeforce ---A. Milking cows

A. Milking cows

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).

Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.

Input

The first line contains an integer n (1 ≤ n ≤ 200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.

Output

Print a single integer, the minimum amount of lost milk.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the%I64d specifier.

Sample test(s)

input

4
0 0 1 0

output

1

input

5
1 0 1 0 1

output

3

 思路：先将脸朝左的牛按顺序从右到左挤，再把脸朝右的牛按顺序从左向右挤，这样才能让损失量最少;

AC Code:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std; 
#define N 200005 
long long int ans, cow[N], lose[N];  
int main(int argc, char const *argv[]) 
{
    int n; 
    /* freopen("in.c", "r", stdin); */ 
    while(cin >> n)
    {
        ans = 0; 
        memset(lose, 0, sizeof(lose)); 
        for(int i = 1;i <= n;i ++)
        {
            cin >> cow[i]; 
            lose[i] = lose[i-1] + cow[i]; 
            if(cow[i] == 0)
                ans += lose[i];  
        }
        cout << ans << endl; 
    }
    return 0; 
}