CodeForces - 598C（高精度几何 排序然后枚举）

Posted 2020-12-25 yinbiao

传送门：

http://codeforces.com/problemset/problem/598/C

Nearest vectors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.

Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 and π. For example, opposite directions vectors have angle equals to π.

Input

First line of the input contains a single integer n (2?≤?n?≤?100?000) — the number of vectors.

The i-th of the following n lines contains two integers x_i and y_i (|x|,?|y|?≤?10?000,?x²?+?y²?>?0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed that no two vectors in the input share the same direction (but they still can have opposite directions).

Output

Print two integer numbers a and b (a?≠?b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any.

Examples

Input

Copy

4
-1 0
0 -1
1 0
1 1

Output

Copy

3 4

Input

Copy

6
-1 0
0 -1
1 0
1 1
-4 -5
-4 -6

Output

Copy

6 5

分析：
题意：
给你n个点，问你这些点和原点构成的n个向量中，哪两个向量构成的角最小
做法：
用atan2函数把所有向量与x轴正半轴的夹角求出，排序，俩俩间比较差值即可。
需要注意的地方：
1.使用long double
2.两两比较的时候，一开始要取最大差值，就是第一个和最好一共的差值
才能比较出最小的嘛
code：

#include<bits/stdc++.h>
using namespace std;
typedef long double LB;
#define max_v 100105
#define PI acos(-1)
#define double long double // 使用long double
const int MAXN = 1e5 + 111;
struct node
{
    double v;
    int index;
    bool operator < (const node& t)const//按照v升序
    {
        return v < t.v;
    }
} p[MAXN];
bool cmp(node a,node b)
{
    return a.v<b.v;
}
int main()
{
    int n;
    scanf("%d",&n);
    int x,y;
    for(int i=1; i<=n; ++i)
    {

        scanf("%d %d",&x,&y);
        p[i].v=atan2(y,x);
        p[i].index=i;
    }
    sort(p+1,p+n+1);
    int ans1=p[1].index;
    int ans2=p[n].index;
    double v=p[1].v+2*PI-p[n].v;
    for(int i=1; i<n; ++i)
    {
        double temp=p[i+1].v-p[i].v;
        if(temp<v)
        {
            v=temp;
            ans1=p[i].index;
            ans2=p[i+1].index;
        }
    }
    printf("%d %d
",ans1,ans2);
    return 0;
}