ural 2064. Caterpillars

2064. Caterpillars

Time limit: 3.0 second
Memory limit: 64 MB

Young gardener didn’t visit his garden for a long time, and now it’s not very pleasant there:  ncaterpillars have appeared on the ground.

Kirill decided to use this opportunity to have some fun and organized a competition — "caterpillar crawl-race."

At Kirill’s command all caterpillars start crawling from the ground to the top of a tree. But they get tired pretty fast. After crawling  t i cm  i-th caterpillar needs to rest for  t i minutes. During that time it slides down a bit. Crawling speed of a caterpillar is 1 cm/minute, sliding speed — also 1 cm/minute.

Kirill is very much interested to find out how high on the tree is the leading caterpillar at different moments in time.

Input

First line contains one integer  n — the number of caterpillars (1 ≤  n ≤ 106).

Second line contains  n integers  t i — characteristics of caterpillars (1 ≤  t i ≤ 109).

In the third line there is a number  q — number of moments in time, which Kirill finds interesting (1 ≤  q ≤ 106).

Remaining  q lines contain one query from Kirill each. A query is described by  x i — number of minutes since the start of the competition (1 ≤  x i ≤ 106).

Output

For every query print in a separate line one integer, that describes how high is the highest caterpillar at the given moment of time.

Sample

input	output
4 1 3 2 1 12 1 2 3 4 5 6 7 8 9 10 11 12	1 2 3 2 1 2 1 2 3 2 1 0

Problem Author: Nikita Sivukhin (prepared by Alexey Danilyuk, Nikita Sivukhin)
Problem Source: Ural Regional School Programming Contest 2015

Tags: none   ( hide tags for unsolved problems )

Difficulty: 559

 

题意：有n只蜗牛，对于第i只蜗牛，有性质ti，这只蜗牛往上爬ti秒，有下降ti秒，上升下降速度都是1cm/s，问第x秒爬得最高的蜗牛多高。

分析：首先画出蜗牛们的高度的函数，就可以看到它们是周期的类似三角形的函数。

如果我们只关注最高点，那么第x秒的答案是

max{a[i] - abs(x - i)}

其中a[i]表示在第i秒这个点的最高点最大是多少。

把那个abs拆开，分成x<i,x>i讨论

发现当x>i时

a[i]+i - x

当x<i时

a[i]-i  + x

所以用是不是有点像单调队列？

其实我们只用记录最大值即可。

 

 1 /**
 2 Create By yzx - stupidboy
 3 */
 4 #include <cstdio>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <cmath>
 8 #include <deque>
 9 #include <vector>
10 #include <queue>
11 #include <iostream>
12 #include <algorithm>
13 #include <map>
14 #include <set>
15 #include <ctime>
16 #include <iomanip>
17 using namespace std;
18 typedef long long LL;
19 typedef double DB;
20 #define MIT (2147483647)
21 #define INF (1000000001)
22 #define MLL (1000000000000000001LL)
23 #define sz(x) ((int) (x).size())
24 #define clr(x, y) memset(x, y, sizeof(x))
25 #define puf push_front
26 #define pub push_back
27 #define pof pop_front
28 #define pob pop_back
29 #define mk make_pair
30 
31 inline int getInt()
32 {
33     int ret = 0;
34     char ch = ' ';
35     bool flag = 0;
36     while(!(ch >= '0' && ch <= '9'))
37     {
38         if(ch == '-') flag ^= 1;
39         ch = getchar();
40     }
41     while(ch >= '0' && ch <= '9')
42     {
43         ret = ret * 10 + ch - '0';
44         ch = getchar();
45     }
46     return flag ? -ret : ret;
47 }
48 
49 const int N = 1000010;
50 int n, m, cnt[N];
51 int ans[N];
52 
53 inline void input()
54 {
55     for(cin >> n; n--; )
56     {
57         int x, t;
58         cin >> x;
59         for(t = x; t < N; t += 2 * x)
60             cnt[t] = max(cnt[t], x);
61         cnt[N - 1] = max(cnt[N - 1], x - (t - (N - 1)));
62     }
63     cin >> m;
64 }
65 
66 inline void solve()
67 {
68     for(int i = 1, now = -INF; i < N; i++)
69     {
70         now = max(now, cnt[i] + i);
71         ans[i] = max(ans[i], now - i);
72     }
73     for(int i = N - 1, now = -INF; i > 0; i--)
74     {
75         now = max(now, cnt[i] - i);
76         ans[i] = max(ans[i], now + i);
77     }
78     
79     while(m--)
80     {
81         int x;
82         cin >> x;
83         cout << ans[x] << "\n";
84     }
85 }
86 
87 int main()
88 {
89     ios::sync_with_stdio(0);
90     input();
91     solve();
92     return 0;
93 }

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