题目
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".
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.
Sample Input
Input
5
3 10 8 6 11
4
1
10
3
11Output
0
4
1
5Hint
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.
题解
多次查询小于等于 m
的数据的个数
查询次数较多,直接查找会超时
应该使用二分查找
使用 upper_bound()
可以直接获得答案
upper_bound 和 lower_bound
upper_bound(a,a+n,e)
返回 a ~ a+n
区域内大于 e
的第一个内存地址
用其减去首地址即可得到下标
则可以直接获取大于 e
的第一个数据的下标
如果从 0
开始的话,恰好多的一位补到从 0
开始错的一位上
lower_bound(a,a+n,e)
返回 a ~ a+n
区域大于等于 e
的第一个内存地址
如果 e
不存在,则 lower_bound()
和 upper_bound()
的结果应该是一样的
如果有 e
存在,则 lower_bound()
会返回第一个 e
的地址
而 upper_bound()
会返回最后一个 e
的下一个地址
代码
/* By:OhYee Github:OhYee Blog:http://www.oyohyee.com/ Email:oyohyee@oyohyee.com かしこいかわいい? エリーチカ! 要写出来Хорошо的代码哦~ */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <vector> #include <list> #include <queue> #include <stack> #include <map> #include <set> #include <functional> #include <bitset> #include <iomanip> using namespace std; const int maxn = 100005; int a[maxn]; bool Do() { int n; if(!(cin >> n)) return false; for(int i = 0;i < n;i++) cin >> a[i]; sort(a,a + n); int q; cin >> q; for(int i = 0;i < q;i++) { int t; cin >> t; cout << upper_bound(a,a + n,t) - a << endl; } return true; } int main() { cin.tie(0); cin.sync_with_stdio(false); while(Do()); return 0; }