CodeForces - 427A (警察和罪犯 思维题)

Police Recruits

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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 Input

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

Hint

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.

Source

Codeforces Round #244 (Div. 2)

题意：n个数，-1代表一个罪犯，正数代表有多少警察，每个警察可以调查一个罪犯，罪犯和警察按照顺序依次出现，

　　　如果罪犯出现时警察的个数为0，未被调查的罪犯数+1，输出未被调查的罪犯数。

题解：代码中a代表警察，b代表罪犯，ans代表未被调查的罪犯。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
    int i,n,ans,a,b,data;
    scanf("%d",&n);
    a=0,b=0,ans=0;
    for(i=0;i<n;i++)
    {
        scanf("%d",&data);
        if(data>0)
        {
            a+=data;
        }
        if(data<0)
        {
            b-=data;
            if(b>a)
            {
                ans++;
                b--;
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}