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• 文章目录一、Police Recruits总结 一、Police Recruits 本题链接：Police Recruits 题目： A. Police Recruits time limit per test1 second memory limit per test256 megabytes inputstandard input ...


文章目录
一、Police Recruits总结

一、Police Recruits
本题链接：Police Recruits
题目： A. Police Recruits time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard 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 ≤ 1e5), 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 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.
本博客给出本题截图：
题意：输入-1证明产生了一个犯罪，输入其他正整数(小于10)代表招募警察的数量，每个警察在处理完一个犯罪后就会退休，且初始警察数为0，问有多少起犯罪是没有警察出来的
AC代码
#include <iostream>

using namespace std;

int main()
{
int n;
cin >> n;

int res = 0, cnt = 0;
for (int i = 0; i < n; i ++ )
{
int a;
cin >> a;
if (a == -1)
{
if (cnt == 0) res ++;
else cnt --;
}
else cnt += a;
}

cout << res << endl;

return 0;
}


总结
水题，不解释
展开全文
• UNT-Police
• Police Patrol 题目描述： 题目大概讲述的是，现在要在x轴上建一座警察局，x轴上有n个罪犯，现在警局有一辆最多能够装载m个人的巡逻车，然后题目让我们求这辆巡逻车把所有罪犯运回警察局需要行驶的最短距离是多少...
                                         Police Patrol
题目描述：
题目大概讲述的是，现在要在x轴上建一座警察局，x轴上有n个罪犯，现在警局有一辆最多能够装载m个人的巡逻车，然后题目让我们求这辆巡逻车把所有罪犯运回警察局需要行驶的最短距离是多少。
题目分析：
这个题目警察局的位置不同会影响到答案的不一样，因此我们要尽量把警察局建在一个合适的位置，使得巡逻车所行驶的无用距离最短(所谓的无用距离是指巡逻车从警察局到达第一个罪犯的距离)，如果警察局在所有罪犯的左边或者右边的话，我们会发现无用距离是很大的，而如果警察局在罪犯的中间的话，那么无用距离就会大大减小。
其实警察局的位置不用枚举出来，我们假设警察局在罪犯中间的某个位置，最左边的罪犯到最右边的罪犯的距离是无论警察局位置在哪里都不会改变的，这个距离的两倍就是巡逻车从警察局到左边搭载u个罪犯(u<=m)以及右边搭载v个罪犯(v<=m)需要行驶的距离，根据贪心策略，巡逻车每次从警察局出发无论向左边还是右边，尽可能运输多一些的罪犯，模拟这个过程，直到所有罪犯都被运输完后，以上每次运输罪犯需要行驶的距离的总和便是答案。
代码：
#include <iostream>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <stdlib.h>
#include <cmath>
#include <math.h>
#include <algorithm>
#include <cstring>
#include <string>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <deque>
#define reg register
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define lowbit(x) (x&(-x))
using namespace std;
const int Maxn=1e6+5;
ll x[Maxn];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (reg int i=1;i<=n;i++) scanf("%lld",&x[i]);
int l=1,r=n;
ll ans=0;
while (l<r)
{
ans+=(x[r]-x[l])<<1;
l+=m;
r-=m;
}
printf("%lld\n",ans);
return 0;
}

展开全文
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• G - Police Recruits 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 ...
G - Police Recruits 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 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.
#include <iostream>

using namespace std;
int a[22222222];
int main()
{
int n,i;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
int x=0,y=0;
for(i=1;i<=n;i++)
{
if(a[i]>0)
{
x+=a[i];
}
else
{
if(x>0)
{
x--;
}
else
y++;
}
}
cout<<y;
return 0;
}

展开全文
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...