HDU4367(线段相交与斐波那契数列)

题目：The war of virtual world

 

题意：在平面内给n个点的坐标，n小于等于200，在这n个点中先选两个点a，b，有(n-1)*n/2种，对于每一种情况，分别求出Ki，Ki等于选定的a，b直线

与剩下的点的交点数，求表达式：

的值。

 

#include <string.h>#include <algorithm>#include <math.h>#include <stdio.h>using namespace std;typedef long long LL;const int N=210;const int M=41000;const int MOD=1000000007;const double eps=1e-8;const double PI=acos(-1.0);const double PI2=PI*2;struct Point{    double x,y;    LL index;    double angle;    inline void input()    {        scanf("%lf%lf",&x,&y);    }}point[N],temp[N];int n;bool visit[M];LL fib[M];LL fk[M];int num[N][N];int left[N][N];int right[N][N];inline Point operator-(const Point &a, const Point &b){    Point c;    c.x=a.x-b.x;    c.y=a.y-b.y;    return c;}inline double operator*(const Point &a, const Point &b){    return a.x*b.y-a.y*b.x;}inline bool operator==(const Point &a, const Point &b){    return a.x==b.x&&a.y==b.y;}inline int cross(const Point &a, const Point &b, const Point &o){    return (a-o)*(b-o)>0? 1:-1;}inline int cross(const int &a, const int &b, const int &o){    return cross(point[a],point[b],point[o]);}inline double positiveAtan(const Point &a, const Point &o){    double res=atan2(a.y-o.y,a.x-o.x);    if(res<0)        res+=PI2;    return res;}bool operator<(const Point &a, const Point &b){    return a.angle<b.angle;}int abs1(int x){    return x<0? -x:x;}int getAngleNumber(int a,int b,int c){    if(point[c].y<point[b].y&&point[a].y>=point[b].y)        return n-abs1(right[b][c]-right[b][a]+2)+3;    return abs1(right[b][a]-right[b][c])+2;}int getTriangleNumber(int a, int b, int c){    return n-left[a][b]-left[b][c]-left[c][a]+getAngleNumber(a,b,c)+getAngleNumber(b,c,a)+getAngleNumber(c,a,b)-6;}LL quick_mod(LL a,LL b){    LL ans=1;    a%=MOD;    while(b)    {        if(b&1)        {            ans=ans*a%MOD;            b--;        }        b>>=1;        a=a*a%MOD;    }    return ans;}LL solve(int x){    if(visit[x])        return fk[x];    visit[x]=true;    fib[0]=x;    fib[1]=x;    for(int i=2;i<=x;++i)        fib[i]=(fib[i-1]*fib[i-2])%MOD;    return fk[x]=fib[x]+1;}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;++i)        {            point[i].input();            temp[i]=point[i];            temp[i].index=i;        }        memset(left,0,sizeof(left));        memset(right,0,sizeof(right));        for(int i=0;i<n;++i)        {            for(int j=i+1;j<n;++j)            {                for(int k=0;k<n;++k)                {                    if(k!=i&&k!=j)                    {                        if(cross(k,j,i)<0)                            ++left[i][j];                        else if(cross(k,i,j)<0)                            ++left[j][i];                    }                }            }            for(int j=0;j<n;++j)            {                if(temp[j].index==i)                    temp[j].angle=-1e100;                else                    temp[j].angle=positiveAtan(temp[j],point[i]);            }            sort(temp,temp+n);            int cnt=0;            for(int j=0;j<n;++j)                right[i][temp[j].index]=++cnt;        }        memset(num,0,sizeof(num));        for(int i=0;i<n;++i)        {            for(int j=i+1;j<n;++j)            {                for(int k=0;k<n;++k)                {                    if(k==i||k==j) continue;                    if(cross(point[k], point[j], point[i]) < 0)                        num[i][j]+=getAngleNumber(j,k,i)-getTriangleNumber(i,j,k);                }            }        }        LL ans=1;        for(int i=0;i<n;++i)            for(int j=i+1;j<n;++j)                ans=(ans*solve(num[i][j]))%MOD;        printf("%I64d\n", ans);    }    return 0;}