CodeForces 427A Police Recruits

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本文介绍了一种简单的警力调度模拟算法，通过输入一系列事件（犯罪发生或警员招募），模拟警力调度过程，并计算无法得到及时处理的犯罪数量。

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.

Examples

input

3
-1 -1 1

output

input

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

output

input

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

output

Note

Lets consider the second example:

Firstly one person is hired.
Then crime appears, the last hired person will investigate this crime.
One more person is hired.
One more crime appears, the last hired person will investigate this crime.
Crime appears. There is no free policeman at the time, so this crime will go untreated.
One more person is hired.
One more person is hired.
One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.

这道题比较水，虽然题目字数比较多，但是直接看答案和NOTE就可以理解
题意大致是这样的：
如果说负数表示小偷，正数表示警察，一开始警察和小偷数都是0，如果出现了小偷但是没有警察，则发生了一个案件，如果有警察了，再出现小偷，小偷就会被抓住。输出案件数量。（这样理解比较容易）

AC

#include <cstdio>

int main()
{
    int n;
    int m;
    int a=0,b=0;//a=thief,b=police
    int sum=0;//案件数
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d",&m);
        if(m>0&&b>=0)b+=m;//m>0，为警察
        if(m<0&&b==0)sum-=m;//m<0,为小偷，如果此时没有警察，此时发生案件
        else if(m<0&&b>0)//m<0,有小偷，但是警察数目不为0
        {
            a=b+m;//没有被抓住的小偷数目，则发生案件
            b=b+m;//剩余警察
            if(a<0)
                sum-=a;
        }
    }
    printf("%d",sum);
}