Interesting drink

题目链接：

http://codeforces.com/contest/706/problem/B

Description

```
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.

Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".

</big>

##Input

<big>

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy's favourite drink.

The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.

Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.

</big>

##Output

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Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.

</big>

##Examples

<big>

input

5

3 10 8 6 11

4

1

10

3

11

output

0

4

1

5

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##Source

<big>

Codeforces Round #367 (Div. 2)

</big>

<br/>

##题意：

<big>

给出一种饮料在不同店的价格，给出每天的消费额度，求每天可以选择多少个店.

</big>

<br/>

##题解：

<big>

预处理一个前缀和即可.

</big>

<br/>

##代码：

``` cpp

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <map>

#include <set>

#include <stack>

#include <vector>

#include <list>

#define LL long long

#define eps 1e-8

#define maxn 101000

#define mod 100000007

#define inf 0x3f3f3f3f

#define mid(a,b) ((a+b)>>1)

#define IN freopen("in.txt","r",stdin);

using namespace std;

int cnt[maxn];

int main(int argc, char const *argv[])

{

    //IN;

    int n;

    while(scanf("%d", &n) != EOF)

    {

        memset(cnt, 0, sizeof(cnt));

        for(int i=1; i<=n; i++) {

            int x; cin >> x;

            cnt[x]++;

        }

        for(int i=1; i<maxn; i++)

            cnt[i] += cnt[i-1];

        int q; cin >> q;

        while(q--) {

            int x; cin >> x;

            if(x >= maxn-1) printf("%d\n", cnt[maxn-1]);

            else printf("%d\n", cnt[x]);

        }

    }

    return 0;

}