codeforces 257C解题报告

这破题，开始不知道是算法挂了，和正确答案的差距很小，还以为是精度不够。。题意：有个初始的坐标(0, 0) 然后有n个人的坐标(xi, yi) 你的任务就是求出最小的扇形的角度，覆盖所有的人，扇形r无限大一开始直接atan2(y, x)求出角度，然后加PI排个序，最大减最小输出，wa到死。这题该先存下角度，然后排个序，在枚举最接近的两个点，假设这两个点之间的扇形区域就是我们求的区域的补，那么只需要这个区域最大，答案就最小了

////  Created by Matrix on 2016-01-22//  Copyright (c) 2015 Matrix. All rights reserved.//////#pragma comment(linker, "/STACK:102400000,102400000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e5 + 10;const int maxv = 1e3 + 10;const double eps = 1e-10;const double PI = acos(-1);double x[maxn], y[maxn];double angle[maxn];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios::sync_with_stdio(0);    int n;    while(cin >> n) {        for(int i = 1; i <= n; i++) {            cin >> x[i] >> y[i];        }        for(int i = 1; i <= n; i++) {            angle[i] = atan2(y[i], x[i]);        }        sort(angle + 1, angle + 1 + n);        double res = 2 * PI;        angle[n+1] = angle[1] + 2.0 * PI;        for(int i = 1; i <= n; i++) {            res = min(res, 2 * PI - fabs(angle[i + 1] - angle[i]));        }//      printf("%f\n", res);        cout << setprecision(10) << res / acos(-1) * 180.0 << endl;    }    return 0;}