URAL 1787. Turn for MEGA

1787. Turn for MEGA

Time limit: 1.0 second
Memory limit: 64 MB

A traffic light at the turn for the “MEGA” shopping center from the Novomoskovskiy highway works in such a way that k cars are able to take a turn in one minute. At weekends all the residents of the city drive to the mall to take a shopping, which results in a huge traffic jam at the turn. Administration of the mall ordered to install a camera at the nearby bridge, which is able to calculate the number of cars approaching this turn from the city. The observation started n minutes ago. You should use the data from the camera to determine the number of cars currently standing in the traffic jam.

Input

The first line contains integers k and n (1 ≤ k, n ≤ 100), which are the number of cars that can take a turn to “MEGA” in one minute and the number of minutes passed from the beginning of observation. The second line contains space-separated integers a1, …, an (0 ≤ ai ≤ 100), where aiis the number of cars that approached the turn during the i-th minute. The observation started at morning, when there were no cars at the turn.

Output

Output the number of cars currently standing in the traffic jam.

Samples

input	output
5 3 6 7 2	0
5 3 20 0 0	5

题的大概意思就是在一个路口，每分钟可以有k辆车转头，n分钟内分别有ai辆车通过，输出有多少车滞留。

刚开始想的太简单了，直接用了个sum把ai全加起来和k*n比较，后来发现下一分钟接近路口的车不能出现在上一分钟里。

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
    int k,n,sum;
    int a[1001];
    while(cin>>k>>n)
    {
        sum=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(sum+a[i]>k)
                sum+=a[i]-k;
            else
                sum=0;
        }
        cout<<sum<<endl;
    }
    return 0;
}