Codeforces - 1166C - A Tale of Two Lands

Codeforces - 1166C - A Tale of Two Lands

地址

http://codeforces.com/contest/1166/problem/C

原文地址

https://www.lucien.ink/archives/432

题目

The legend of the foundation of Vectorland talks of two integers $x$ and $y$. Centuries ago, the array king placed two markers at points $|x|$ and $|y|$ on the number line and conquered all the land in between (including the endpoints), which he declared to be Arrayland. Many years later, the vector king placed markers at points $|x-y|$ and $|x+y|$ and conquered all the land in between (including the endpoints), which he declared to be Vectorland. He did so in such a way that the land of Arrayland was completely inside (including the endpoints) the land of Vectorland.

Here $|z|$ denotes the absolute value of $z$.

Now, Jose is stuck on a question of his history exam: “What are the values of $x$ and $y$?” Jose doesn’t know the answer, but he believes he has narrowed the possible answers down to $n$ integers $a_1,a_2,\dots ,a_n$. Now, he wants to know the number of unordered pairs formed by two different elements from these $n$ integers such that the legend could be true if $x$ and $y$ were equal to these two values. Note that it is possible that Jose is wrong, and that no pairs could possibly make the legend true.

题意

给你 $n$ 个数，问其中有多少对 x y 满足 $min(|x-y|,|x+y|)\le min(|x|,|y|)$ 且 $max(|x|,|y|)\le max(|x-y|,|x+y|)$

题解

分类讨论一下，不妨设 $|x|\le |y|$

1. $0<x,y$

$$y-x\le x\le y\le x+y\Rightarrow x\le y\le 2x$$

2. $x<0$, $y>0$

$$x+y\le -x\le y\le y-x\Rightarrow -x\le y\le -2x$$

3. $x>0$, $y<0$

$$-y-x\le x\le -y\le x-y\Rightarrow x\le -y\le 2x$$

4. $0>x,y$

$$\begin{gathered} x-y\le -x\le -y\le -x-y \\ \Rightarrow 2x\le y\le x \\ \Rightarrow |x|\le |y|\le 2|x| \end{gathered}$$

5. 只有当 $x$ 和 $y$ 同时为 $0$ 时才满足不等式

即：

找出有多少对 x y 满足 $|x|\le |y|\le 2|x|$ ，lower_bound 一下 upper_bound 一下，容斥一下即可。

代码

https://pasteme.cn/8175

#include 
long long ans, n;
std::vector<int> vec;
std::map<int, int> cnt;
int main() {
    scanf("%d", &n);
    for (int i = 0, buf; i < n; i++) {
        scanf("%d", &buf);
        vec.push_back(abs(buf));
		cnt[abs(buf)]++;
    }
	std::sort(vec.begin(), vec.end());
    for (auto each : vec) {
        auto l = std::lower_bound(vec.begin(), vec.end(), each);
        auto r = std::upper_bound(vec.begin(), vec.end(), each * 2);
        ans += r - l;
    }
    for (auto pair : cnt) ans -= pair.second * pair.second - pair.second * (pair.second - 1) / 2;
    printf("%lld\n", ans);
    return 0;
}