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 ti cm i-th caterpillar needs to rest for ti 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 ti — characteristics of caterpillars (1 ≤ ti ≤ 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 xi — number of minutes since the start of the competition (1 ≤ xi ≤ 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。

 

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define mk make_pair

 inline int getInt()
 {
     int ret = ;
     char ch = ' ';
     bool flag = ;
     while(!(ch >= '' && ch <= ''))
     {
         if(ch == '-') flag ^= ;
         ch = getchar();
     }
     while(ch >= '' && ch <= '')
     {
         ret = ret *  + ch - '';
         ch = getchar();
     }
     return flag ? -ret : ret;
 }

 const int N = ;
 int n, m, cnt[N];
 int ans[N];

 inline void input()
 {
     for(cin >> n; n--; )
     {
         int x, t;
         cin >> x;
         for(t = x; t < N; t +=  * x)
             cnt[t] = max(cnt[t], x);
         cnt[N - ] = max(cnt[N - ], x - (t - (N - )));
     }
     cin >> m;
 }

 inline void solve()
 {
     for(int i = , now = -INF; i < N; i++)
     {
         now = max(now, cnt[i] + i);
         ans[i] = max(ans[i], now - i);
     }
     for(int i = N - , now = -INF; i > ; i--)
     {
         now = max(now, cnt[i] - i);
         ans[i] = max(ans[i], now + i);
     }

     while(m--)
     {
         int x;
         cin >> x;
         cout << ans[x] << "\n";
     }
 }

 int main()
 {
     ios::sync_with_stdio();
     input();
     solve();
     return ;
 }