题解 | #牛可乐和魔法封印#

居然没人写题解，就让我来勇闯无人区！
这个题目好像就类似于高中的VB题，上代码

int findlowerbound(int target, int numsize) {
	int l = 0;
	int r = numsize - 1;
	while (l <= r)
	{
		int m = (l + r) / 2;
		if (num[m] < target)
		{
			l = m + 1;
		}
		else {
			r = m - 1;
		}
	}
	return l;
}

我们高中的时候好像是用i代替l，j代替r，然后用一个叫做key的变量代替target（死去的记忆开始攻击我）
总之完整的代码是这样的

#include<iostream>
using namespace std;
#define MAXN 100010
int _;
int num[MAXN];
int findlowerbound(int target, int numsize) {
	int l = 0;
	int r = numsize - 1;
	while (l <= r)
	{
		int m = (l + r) / 2;
		if (num[m] < target)
		{
			l = m + 1;
		}
		else {
			r = m - 1;
		}
	}
	return l;
}
int findUpperbound(int target, int numsize) {
	int l = 0;
	int r = numsize - 1;
	while (l <= r)
	{
		int m = (l + r) / 2;
		if (num[m] <= target)
		{
			l = m + 1;
		}
		else {
			r = m - 1;
		}
	}
	return r;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int _;
	cin >> _;
	for (int i = 0;i < _;i++) {
		cin >> num[i];
	}
	int n;
	cin >> n;
	for (int i = 0;i < n;i++) {
		int lowertarget, hightarget;
		cin >> lowertarget >> hightarget;
		int result = findUpperbound(hightarget, _) - findlowerbound(lowertarget, _) + 1;
		cout << result << endl;
	}
}