codeforces難度，codeforces 598C C. Nearest vectors(極角排序)

題目鏈接：

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

codeforces難度，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

題意：找到兩個向量間夾角最小的那兩個向量的位置;
思路：直接暴力絕對絕對絕對會超時,所以要先按極角排序,排完后再找兩個相鄰的向量夾角最小的那對,一開始自己用余弦定理求角發現精度不夠,看網上說用long double ,改成long double 后還是被test104和test105卡死了,所以換成atan2函數最后才過,看來余弦定理求還是精度不行;
AC代碼：

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+20;
const long double PI=acos(-1.0);
struct node
{int num;long double x,y;long double angle;
};
node point[N];
int cmp(node s1,node s2)
{return s1.angle<s2.angle;
}
int main()
{int n;double xx,yy;scanf("%d",&n);for(int i=1;i<=n;i++){cin>>xx>>yy;//scanf("%lf%lf",&xx,&yy);point[i].x=xx;point[i].y=yy;point[i].num=i;point[i].angle=atan2(yy,xx);//point[i].angle=acos(xx/sqrt(xx*xx+yy*yy));//if(yy<0)point[i].angle=2*PI-point[i].angle;
    }sort(point+1,point+n+1,cmp);int ansa,ansb;long double mmin=90,w;long double x1,y1,x2,y2;for(int i=2;i<=n;i++){x1=point[i].x;y1=point[i].y;x2=point[i-1].x;y2=point[i-1].y;w=atan2(y1,x1)-atan2(y2,x2);if(w<0)w+=2*PI;//acos((x1*x2+y1*y2)/(sqrt(x1*x1+y1*y1)*sqrt(x2*x2+y2*y2)));if(w<=mmin){ansa=point[i-1].num;ansb=point[i].num;mmin=w;}}x1=point[1].x;y1=point[1].y;x2=point[n].x;y2=point[n].y;w=atan2(y1,x1)-atan2(y2,x2);if(w<0)w+=2*PI;//w=acos((x1*x2+y1*y2)/(sqrt(x1*x1+y1*y1)*sqrt(x2*x2+y2*y2)));if(w<mmin){ansa=point[1].num;ansb=point[n].num;mmin=w;}printf("%d %d\n",ansa,ansb);

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