AtCoder Beginner Contest 139F Engines

link

problem

Given \ (n-\) th tuple \ ((X, Y) \) . Initially at the origin \ ((0,0) \) , each can from \ (n-\) picked bins and a then current coordinate \ ((X, Y) \ ) becomes \ ( (+ X-X, the Y + Y) \) , each tuple can only be selected once. Some selected tuple, the tuple such that after moving the maximum Euclidean distance according to the origin. Find the distance.

solution

These \ (n-\) bins and seen \ (n-\) vectors. Moves to follow the parallelogram. Therefore, the smaller the angle between two vectors, the vector sum formation, and the greater the length of the mold. Therefore, these vectors sorted polar angle. Selection vector must be an interval. Enumeration about the endpoint, you can find the maximum value.

code

//@Author: wxyww
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
#include<map>
#include<string>
using namespace std;
typedef long long ll;
#define pi pair<double,double>
const int N = 110;
ll read() {
    ll x = 0,f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0',c = getchar();}
    return x * f;
}
pi a[N];
bool cmp(const pi &A,const pi &B) {
    return atan2(A.second , A.first) < atan2(B.second , B.first);
}
int main() {
    int n = read();
    for(int i = 0;i < n;++i) a[i].first = read(),a[i].second = read();
    sort(a,a + n,cmp);
    double ans = 0;
    for(int i = 0;i < n;++i) {
        double X = a[i].first,Y = a[i].second;
        ans = max(ans,X * X + Y * Y);
        for(int j = (i + 1) % n;j != i;j = (j + 1) % n) {
            X += a[j].first,Y += a[j].second;
            ans = max(ans,X * X + Y * Y);
        }
    }
    printf("%.10lf",sqrt(ans));
    return 0;
}