Codeforces 706B Interesting drink 【二分】

Interesting drink

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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".

Input

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.

Output

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.

Example

input

5
3 10 8 6 11
4
1
10
3
11

output

0
4
1
5

Note

On the first day, Vasiliy won't be able to buy a drink in any of the shops.

On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.

On the third day, Vasiliy can buy a drink only in the shop number 1.

Finally, on the last day Vasiliy can buy a drink in any shop.

lower_bound(k)寻找  k <= ? 并返回 ，upper_bound(k)寻找 k < ? 并返回。
#include <cmath>
#include <cstdio>
#include <set>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
//#define LOCAL
#define space " "
using namespace std;
//typedef __int64 Int;
//typedef long long Long;
const int INF = 0x3f3f3f3f;
const int MAXN = 100000 + 10;
const double Pi = acos(-1.0);
const double ESP = 1e-5;
int main() {
int n, ar[MAXN], m, a;
while (scanf("%d", &n) != EOF) {
for (int i = 0; i < n; i++) {
scanf("%d", &ar[i]);
}
sort(ar, ar + n); scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &a);
int ans = upper_bound(ar, ar + n, a) - ar;
printf("%d\n", ans);
}
}
return 0;
}