CF1971F Circle Perimeter

合集 - 有意思的题目(7)

1.CF1971F Circle Perimeter2024-06-03

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题目

Given an integer $r$, find the number of lattice points that have a Euclidean distance from $(0,0)$ greater than or equal to $r$ but strictly less than $r+1$.

A lattice point is a point with integer coordinates. The Euclidean distance from $(0,0)$ to the point $(x,y)$ is $\sqrt{x^2+y^2}$.
https://codeforces.com/contest/1971/problem/F

Input

The first line contains a single integer $t$ ($1\le t\le 1000$) — the number of test cases.The only line of each test case contains a single integer $r$ ($1\le r\le {10}^5$).
The sum of $r$ over all test cases does not exceed ${10}^5$.

6
1
2
3
4
5
1984

Output

For each test case, output a single integer — the number of lattice points that have an Euclidean distance $d$ from $(0,0)$ such that $r\le d<r+1$.

8
16
20
24
40
12504

题解

解题思路

注意到数据范围为1e5级别，需要线性时间复杂度，故考虑枚举一个数并快速算出另外一个数
对公式$r\le \sqrt{x^2+y^2}\le r+1$进行变形，可以得知需要枚举x（范围为$(-r,r)$），利用公式求出y从而快速计算，注意边界情况：$x^2+y^2\le (r+1{)}^2-1$

Code

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;
signed main(){
	IOS;
    int _;cin>>_;
    while(_--){
        LL n;cin>>n;
        LL ans=0;
        for(LL i= -n;i<=n;i++){
            LL t1=ceil(sqrtl(n*n-i*i));
            LL t2=sqrtl((n+1)*(n+1)-i*i-1);
            ans+=(t2-t1+1)*2-(t1==0);
        }
        cout<<ans<<endl;
    }
	
	return 0;
}