codeforces_598C. Nearest vectors（极角排序）

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 xi and yi (|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

4
-1 0
0 -1
1 0
1 1

output

3 4

input

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

output

6 5

计算几何的极角排序，用atan2()来排，坑点在long double 。。。

atan2(y,x)计算点（x,y）和x正半轴的夹角（弧度制）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 100010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

struct node
{
    int id;
    long double at;
    friend bool operator <(node a,node b)
    {
        return a.at<b.at;
    }
} p[maxn];

int main()
{
    int n;
    Si(n);
    double x,y;
    for(int i=0; i<n; i++)
    {
        Sd(x),Sd(y);
        p[i].id=i;
        p[i].at=atan2(y,x);
    }
    sort(p,p+n);
    p[n]=p[0];
    long double mi=100*PI;
    int ans1=0,ans2=0;
    for(int i=1; i<=n; i++)
    {
        long double temp=p[i].at-p[i-1].at;
        if(i==n)temp+=2*PI;
        if(mi>temp)
        {
            mi=temp;
            ans1=p[i].id;
            ans2=p[i-1].id;
        }
    }
    printf("%d %d\n",ans1+1,ans2+1);
    return 0;
}