[hdu] 6097 "优秀的黄金分割三分"

来源：互联网发布：出口退税算法编辑：程序博客网时间：2024/06/05 08:39
题解写了这句话，比赛时写了三分。T 到飞起。赛后发现三分写丑了。然后，还是需要快速读入。果然我还是菜呀、

#include"bits/stdc++.h"using namespace std;const double Pi = acos(-1.0);double x1, y1, x2, y2;double rr;typedef long long LL;struct FastIO{    static const int S = 2 * 100;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) exit(0);        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for (; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(LL x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;double get_lens(double arg1){    double x11 = rr * cos(arg1);    double y12 = rr * sin(arg1);    return sqrt((x11 - x1) * (x11 - x1) + (y12 - y1) * (y12 - y1)) + sqrt((x11 - x2) * (x11 - x2) + (y12 - y2) * (y12 - y2));}int main(){    //freopen("data.in", "r", stdin);    //freopen("data1.out", "w", stdout);    int T;    double d = (sqrt(5) - 1) / 2;    T = io.xint();    while(T --){        rr = io.xint();        x1 = io.xint(); y1 = io.xint();        x2 = io.xint(); y2 = io.xint();        double t = x1 * x1 + y1 * y1;        double a = atan2(y1, x1);        double b = atan2(y2, x2);        if(a < 0) a += 2 * Pi;        if(b < 0) b += 2 * Pi;        double l, r;        if(a > b){            r = a; l = b;        }        else{            r = b; l = a;        }        if(r - l > Pi){            r -= 2 * Pi;            swap(l , r);        }        r = (l + r) / 2.0;        double midl = (l * d + r * (1 - d));        double midr = (l * (1 - d) + r * d);        double lans = get_lens(midl);        double rans = get_lens(midr);        for(int i = 1; i <= 30; i ++){            if(lans < rans){                r = midr;                midr = midl;                midl = l * d + r * (1 - d);                rans = lans;                lans = get_lens(midl);            }            else{                l = midl;                midl = midr;                midr = l * (1 - d) + r * d;                lans = rans;                rans = get_lens(midr);            }        }        printf("%.12f\n", get_lens(l));    }    return 0;}
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