Codeforces Round #244 (Div. 2)a Police Recruits

A. Police Recruits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.

Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.

If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.

Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.

Input

The first line of input will contain an integer n (1 ≤ n ≤ 105), the number of events. The next line will contain n space-separated integers.

If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.

Output

Print a single integer, the number of crimes which will go untreated.

Sample test(s)

input

3
-1 -1 1

output

2

input

8
1 -1 1 -1 -1 1 1 1

output

1

input

11
-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1

output

8

Note

Lets consider the second example:

1. Firstly one person is hired.
2. Then crime appears, the last hired person will investigate this crime.
3. One more person is hired.
4. One more crime appears, the last hired person will investigate this crime.
5. Crime appears. There is no free policeman at the time, so this crime will go untreated.
6. One more person is hired.
7. One more person is hired.
8. One more person is hired.

The answer is one, as one crime (on step 5) will go untreated.

首先給出n個整數,正數表示新增多少個警察，-1表示要派出一個警察，當爲-1且沒有警察可派時，結果加1；

思路：模擬即可

#include<cstdio>
int main()
{
    int n,ans,i,d,s;
    while(scanf("%d",&n)!=EOF)
    {
        s=0;
         ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&d);
            if(d>0)
                s+=d;
            else
            {
                if(s+d<0)
                    ans+=1;
                else
                    s=s+d;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}