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  • K-POP Dance 1 1.0.zip

    2021-05-03 21:57:26
    dance 1v0模型动作?(含骨骼绑定)
  • ugg sundance boots

    2009-06-02 11:22:00
    ugg sundance bootsWhen the US I just ugg sundance boots didn’t get it. And at £120 a pop for a basic sandal, or £35 for a sheepskin-lined flip-flop, the new collections corporation place is ...

    ugg sundance boots
    When the US I just ugg sundance boots didn’t get it. And at £120 a pop for a basic sandal, or £35 for a sheepskin-lined flip-flop, the new collections corporation place is unsurprisingly given to the trademark sheepskin boot. These days they come in fuchsia, turquoise and lime-green Deckers Outdoor bought Ugg Australia in 1995, it probably thought it had made a sensible addition to its portfors. bought 350 pairs for her staff and featured the Ultra boot on her Favorite Things show. Ugg went mega, becoming the footwear of choicrs and wannabes.

    Today, in Britain,footwear brands. Ugg was ugg sundance positioned as high-end luxury footwear, best known for its sheepskin boots, favoured by Californian In 2000 Pride of and you can choose from knitted, canvas or sheepskin exteriors a wily PR person sent some to Oprah Winfrey, aware of the power of her endorsement. She loved them so much that she comfort. But many people hate them – let’s call it the Marmite effect. And as they’re not famous for giving you great legs, men are not keen either.

    Ugg opened its a generation of young cheap ugg cardy boots women. Sloaney girls love wearing them with jeans tucked in. Some members of the fashion press wear them for opinions are new season walls, wood and bright lighting, with minimal merchandise displayed on wood and glass shelves. It’s easy to work divided. Teamed with a bomber jacket, miniskirt and black tights, the boots form part of the staple uniform for first British and the queues formed again. I decided to join them.

    The windows Covent ugg mini boots show that Ugg sales were up 57 per cent on last year. (Is this brand recession proof?) Shortly after that debut another shop opened Garden standalone your way around, although a bit narrow – men’s to the left, women’s to the right and children’s at the back. . It’s a classic store, in the Westfield centre, London, last autumn. People queued to get in, and recent figures eye-catching colours let down by a dull display.

    Shopability The shop is Next ugg tall boots for company, rather than cool Neal Street. One window displayed three pairs of frumpy sandals; another, three pairs a mix of brick a great in Covent Garden is ultra-bright sheepskin boots – location, but I was surprised Ugg had chosen a unit with the traffic of Long Acre and brands such as M&S and of the style that has proven staying power. The sandal range http://www.uggslife.com (launched in Australia last September) is another matter. also felt too expensive.

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  • 题目描述: 6.) Proving Afzal Wrong ... Compare the popularity of dance music genres and pop music genres across the dataset using appropiate visualisation/s. Make the assumption that the pop

    题目描述:

    6.) Proving Afzal Wrong

    We have detoured from the original aim of this question for long enough. Compare the popularity of dance music genres and pop music genres across the dataset using appropiate visualisation/s. Make the assumption that the popularity of a genre is defined by the average popularity column entry across all songs in the appropriate genres.

    Hint/s

    • Dance sub-genres can be considered: edm, dance pop, trap music, big room, brostep
    • Pop sub-genres can be considered anything with pop in the name

     

    原始数据集:

    本题我卡在了如何找出Genre列含有’pop‘字段的行,如行0、行2、行3等。然后解决这个题还涉及一些python的常识性小tips,就记录一下。

    搜索了众多python函数后,我还是没有找到可以一键替换的函数,看来只能遍历了。

    import re #正则表达式的包
    import pandas as pd
    
    
    songData['dance_or_pop'] = '' # 新建一列来存储音乐类型
    songData['dance_or_pop'].loc[(songData['Genre'] == 'edm')|(songData['Genre'] == 'brostep')|(songData['Genre'] == 'dance pop')|(songData['Genre'] == 'trap music')|(songData['Genre'] == 'big room')] = 'dance' # 根据题目给舞曲型赋值
    
    pop_song = re.compile('.*pop.*') # 定义正则表达式,即任何含pop的字段
    
    for i in songData['Genre']._stat_axis.values : # 根据行号遍历dataframe
        item = songData.loc[i,'Genre'] # python中如何找到某特定单元格的内容
        if (re.match(pop_song, item) != None): # python中函数返回为空是等于None
            songData.loc[i, 'dance_or_pop'] = 'pop'
    
    songData

    完成后是这个样子:

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  • <div><p>PicoTorrent ... For example (Pop, Dance-Pop) Estradarada - Дискотека Века (2017), MP3, tracks, 320 kbps</code>.</p><p>该提问来源于开源项目:picotorrent/picotorrent</p></div>
  • LeTV_v6.0.1

    2012-10-08 23:16:46
    Normal 标准 Classical 古典 Dance 舞曲 Flat 降音 Folk 民歌 Heavy Metal 重金属 Hip Hop 说唱乐 Jazz 爵士乐 Pop 流行乐 Rock 摇滚乐 Manual 自定义(手动)
  • <p>Why: If we make mistakes in checking the boxes and changes while still in the pop up window, there is a dance of tagging happening in the history of the bug.</p><p>该提问来源于开源项目:...
  • 算法几何 ChaoYeon

    2012-08-27 08:29:32
    Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in he
    Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in her live performances. If you had ever seen her dance, I bet you’d love it.
    I felt stage lighting interesting when I was enjoying Chae Yeon’s performance. We all know that stage lighting instruments are used for the concerts and other performances taking place in live performance venues. They are also used to light the stages. Actually this is a color mixing process. There are two types of color mixing: Additive and Subtractive. Most stages use the former and in this case there are three primary colors: red, green, and blue. In the absence of color, or when no colors are showing, the stage is black. If all three primary colors are showing, the result is white. When red and green combine together, the result is yellow. When red and blue combine together, the result is magenta. When blue and green combine together, the result is cyan. When two same color combine together, the result is still that color.

    We have got the coordinate and the primary color of the figure that each Stage Lighting Instrument sent out. For simplicity’s sake, we can just treat the figure as a circle. Of course we’ll know the radius of each colored circle. Some color may be changed based on the Color Mixed Theory we mentioned above. Can you find the area of each color?

    Input

    The first line consists of an integer T, indicating the number of test cases.
    The first line of each case consists of three integers R, G, B, indicating the number of red circles, green circles and blue circles. The next R + G + B lines, each line consists of three integer x, y, r, indicating the coordinate and the radius. The first R lines descript the red circles, the second G lines descript the green circles and the last B lines descript the blue circles.

    Output

    Output seven floating numbers, they are the area of red, green, blue, white, yellow, magenta and cyan. Please take each number with two factional digits.
    Constraints
    0 < T <= 20
    0 <= R, G, B <= 100
    -100 <= x, y <= 100; 0 <= r <= 100

    Sample Input

    1
    1 1 1
    0 2 3
    2 0 3
    -2 0 3

    Sample Output

    9.28 15.04 15.04 4.92 7.04 7.04 1.28

    Hint

    Gentleman’s Reminder: please make sure that your program won’t output “-0.00”.

    Source

     
    解题思路
    根据容斥原理,S圆1,2交=S圆1,2并-S圆1-S圆2。此原理可推广到两类圆的情况,即:S红,绿交=S红,绿并-S红-S绿。
    也能推广到三种颜色圆的情况,即:S红,绿,蓝交=S红,绿,蓝并-S红-S绿-S蓝+S红,绿并+S红,蓝并+S绿,蓝并。
    所以本题就转化成了求圆的并的问题,需要求:S红、S绿、S蓝、S红,绿并、S红,蓝并、S绿,蓝并、S红,绿,蓝并。
    圆的并的解法:
    将每一个圆与其他所有圆的交点找到,并按照与圆心构成的夹角大小来排序,之后逐一取相邻的两点构成圆弧,循环判断圆弧中点是否在其他圆内,如果是,不计算,如果未被覆盖,计算弓形面积+三角形面积。
    其中,多个三角形面积构成多边形面积,取P(0,0),计算公式:(x1*y2-y1*x2)/2。
    多注意浮点数的处理。
    范例的解法是采用C++实现的。
    优化方案一、改用C语言实现,优化排序的算法,原为C++的sort算法,首先不记录交点,改为记录与圆心夹角,采用插入排序,新增一个处理一个。
    优化效果:NOJ:90ms—>60ms。
    优化方案二、取消排序,以及后面的判断是否被覆盖,初始化为0~2pi,当作区间来处理,在计算出夹角后,每次割掉相交的弧度范围,如果跨越了2pi,则按照两段来处理:Ab~2pi和0~Ae。
    优化效果:NOJ:60ms—>20ms。

    范例程序C++
    /**
     * File Name: ChaeYeon.cpp
     * Created Time:  7/4/2009 5:04:19 PM
     * Author: momodi
     * Description: 
     */
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstdlib>
    #include <algorithm>
    #include <vector>
    using namespace std;
    #define NEXT(v, n) (((v) == (n))? 0 : (v))
    #define SZ(x) ((int) x.size())
    template <class T> bool get_max(T& a, const T &b) {return b > a? a = b, 1: 0;}
    template <class T> bool get_min(T& a, const T &b) {return b < a? a = b, 1: 0;}
    #define SQR(v) ((v) * (v))
    const double eps = 1e-9;
    const double pi = acos(-1.0);
    
    int sgn(const double &a) {
        return (a > eps) - (a < -eps);
    }
    
    const int zx[] = {
        0, 1, 0, -1
    };
    const int zy[] = {
        1, 0, -1, 0
    };
    
    struct P {
        double x, y;
        P(const double &_x, const double &_y)
            :x(_x), y(_y) {}
        P() {}
        bool operator == (const P &a) const {
            return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;
        }
        P operator + (const P &a) const {
            return P(x + a.x, y + a.y);
        }
        P operator - (const P &a) const {
            return P(x - a.x, y - a.y);
        }
        P operator * (const double &a) const {
            return P(x * a, y * a);
        }
        P operator / (const double &a) const {
            return P(x / a, y / a);
        }
        P trunc(double a) const {
            a /= sqrt(SQR(x) + SQR(y));
            return P(x * a, y * a);
        }
        P turn_left() const {
            return P(-y, x);
        }
        P turn_right() const {
            return P(y, -x);
        }
        const P& input() {
            scanf("%lf %lf", &x, &y);
            return *this;
        }
        const P& output() const {
            printf("P: %.12lf %.12lf\n", x, y);
            return *this;
        }
    };
    
    double dist2(const P &a, const P &b) {
        return SQR(a.x - b.x) + SQR(a.y - b.y);
    }
    double dist(const P &a, const P &b) {
        return sqrt(SQR(a.x - b.x) + SQR(a.y - b.y));
    }
    double cross(const P &a, const P &b, const P &c) {
        return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
    }
    double dmul(const P &a, const P &b, const P &c) {
        return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);
    }
    
    struct C {
        P mid;
        double r;
        C(const P &_mid, const double &_r)
            :mid(_mid), r(_r) {}
        C() {}
        bool operator == (const C &a) const {
            return mid == a.mid && sgn(r - a.r) == 0;
        }
        bool in(const C &a) const {
            return sgn(r + dist(mid, a.mid) - a.r) <= 0;
        }
        const C &input() {
            mid.input();
            scanf("%lf", &r);
            return *this;
        }
        const C &output() const {
            printf("P: %.12lf %.12lf R: %.12lf\n", mid.x, mid.y, r);
        }
    };
    
    double cal_angle(const C &c, const P &a, const P &b) {
        double k = dmul(c.mid, a, b) / SQR(c.r);
        get_min(k, 1.0);
        get_max(k, -1.0);
        return acos(k);
    }
    
    double cal_area(const C &c, const P &a, const P &b) {
        return SQR(c.r) * cal_angle(c, a, b) / 2 - cross(c.mid, a, b) / 2;
    }
    
    struct cmp {
        P mid;
        cmp(const P &_mid)
            :mid(_mid) {}
        bool operator () (const P &a, const P &b) {
            return atan2(a.y - mid.y, a.x - mid.x) < atan2(b.y - mid.y, b.x - mid.x);
        }
    };
    
    bool circles_intersection(const C &a, const C &b, P &c1, P &c2) {
        double dd = dist(a.mid, b.mid);
        if (sgn(dd - (a.r + b.r)) >= 0) {
            return false;
        }
        double l = (dd + (SQR(a.r) - SQR(b.r)) / dd) / 2;
        double h = sqrt(SQR(a.r) - SQR(l));
        c1 = a.mid + (b.mid - a.mid).trunc(l) + (b.mid - a.mid).turn_left().trunc(h);
        c2 = a.mid + (b.mid - a.mid).trunc(l) + (b.mid - a.mid).turn_right().trunc(h);
        return true;
    }
    
    bool cover(const C &c, const P &a, const P &b, const vector<C> &cir) {
        P p = c.mid + ((a + b) / 2 - c.mid).trunc(c.r);
        for (vector<C>::const_iterator it = cir.begin(); it != cir.end(); ++it) {
            if (sgn(dist2(p, it->mid) - SQR(it->r)) < 0) {
                return true;
            }
        }
        return false;
    }
    
    double cal_area(const vector<C> &in) {
        vector<C> cir;
        for (int i = 0; i < SZ(in); ++i) {
            if (sgn(in[i].r) == 0) {
                continue;
            }
            bool flag = false;
            for (int j = i + 1; j < SZ(in); ++j) {
                if (in[i] == in[j]) {
                    flag = true;
                    break;
                }
            }
            if (flag) {
                continue;
            }
            for (j = 0; j < SZ(in); ++j) {
                if (!(in[i] == in[j]) && in[i].in(in[j])) {
                    flag = true;
                    break;
                }
            }
            if (flag) {
                continue;
            }
            cir.push_back(in[i]);
        }
        vector<vector<P> > point_on_circle(SZ(cir));
        for (i = 0; i < SZ(cir); ++i) {
            for (int z = 0; z < 4; ++z) {
                point_on_circle[i].push_back(cir[i].mid + P(zx[z], zy[z]).trunc(cir[i].r));
            }
        }
        for (i = 0; i < SZ(cir); ++i) {
            for (int j = i + 1; j < SZ(cir); ++j) {
                P a, b;
                if (circles_intersection(cir[i], cir[j], a, b)) {
                    point_on_circle[i].push_back(a);
                    point_on_circle[i].push_back(b);
                    point_on_circle[j].push_back(a);
                    point_on_circle[j].push_back(b);
                }
            }
        }
        for (i = 0; i < SZ(cir); ++i) {
            sort(point_on_circle[i].begin(), point_on_circle[i].end(), cmp(cir[i].mid));
            point_on_circle[i].erase(unique(point_on_circle[i].begin(), point_on_circle[i].end()), point_on_circle[i].end());
        }
        double ans = 0;
        for (i = 0; i < SZ(cir); ++i) {
            for (int j = 0; j < SZ(point_on_circle[i]); ++j) {
                const P &a = point_on_circle[i][j];
                const P &b = point_on_circle[i][NEXT(j + 1, SZ(point_on_circle[i]))];
                if (!cover(cir[i], a, b, cir)) {
                    ans += cross(P(0, 0), a, b) / 2;
                    ans += cal_area(cir[i], a, b);
                }
            }
        }
        return ans;
    }
    double mabs(double a) {
        return a + eps;
    }
    
    int main() {
        int ca;
        scanf("%d", &ca);
        while (ca--) {
            int R, G, B;
            scanf("%d %d %d", &R, &G, &B);
            vector<C> r, g, b, rg, rb, gb, rgb;
            for (int i = 0; i < R; ++i) {
                C tmp;
                tmp.input();
                r.push_back(tmp);
                rg.push_back(tmp);
                rb.push_back(tmp);
                rgb.push_back(tmp);
            }
            for (i = 0; i < G; ++i) {
                C tmp;
                tmp.input();
                g.push_back(tmp);
                rg.push_back(tmp);
                gb.push_back(tmp);
                rgb.push_back(tmp);
            }
            for (i = 0; i < B; ++i) {
                C tmp;
                tmp.input();
                b.push_back(tmp);
                rb.push_back(tmp);
                gb.push_back(tmp);
                rgb.push_back(tmp);
            }
            double rs = cal_area(r);
            double gs = cal_area(g);
            double bs = cal_area(b);
            double rgs = rs + gs - cal_area(rg);
            double rbs = rs + bs - cal_area(rb);
            double gbs = gs + bs - cal_area(gb);
            double rgbs = cal_area(rgb) - rs - gs - bs + rgs + rbs + gbs;
            printf("%.2f %.2f %.2f %.2f %.2f %.2f %.2f\n", 
                    mabs(rs - rgs - rbs + rgbs), 
                    mabs(gs - rgs - gbs + rgbs),
                    mabs(bs - rbs - gbs + rgbs),
                    mabs(rgbs),
                    mabs(rgs - rgbs),
                    mabs(rbs - rgbs),
                    mabs(gbs - rgbs));
        }
        return 0;
    }

    优化方案一
    #include <stdio.h>
    #include "string.h"
    #include "math.h"
    
    #define S(x) (x)*(x)
    const double pi=3.141592653589,e=1e-9;
    
    typedef struct{
    	double x,y,r;
    }C;
    
    int Ne(const double a)
    {
    	return (a>e)-(a<-e);
    }
    
    double dis(const double a,const double b,const double c,const double d)
    {
    	return sqrt(S(a-c)+S(b-d));
    }
    
    void InQue(double p[],int *t)
    {
    	int i,j;
    	double tp;
    	if(p[*t-1]<0)p[*t-1]+=2*pi;
    	if(p[*t-1]>2*pi)p[*t-1]-=2*pi;
    	for (i=0;i<*t-1;i++)
    	{
    		if(p[*t-1]<p[i]){
    			tp=p[*t-1];
    			for (j=*t-1;j>i;j--)
    				p[j]=p[j-1];
    			p[i]=tp;
    		}
    		else if(!Ne(p[*t-1]-p[i])){
    			(*t)--;break;}
    	}
    }
    
    double G_ar(C c[],int k)
    {
    	char cov;
    	int i,j,t=0,q;
    	C Z[300];
    	double p[600],f=0,d,h,a,a1,a2;
    	for (i=0;i<k;i++)
    	{
    		if(c[i].r==0)continue;
    		cov=0;
    		for (j=i+1;j<k;j++)
    		{
    			if(c[i].x==c[j].x && c[i].y==c[j].y && c[i].r==c[j].r){
    				cov=1;break;}
    		}
    		if(cov)continue;
    		for (j=0;j<k;j++)
    		{
    			if (i!=j && c[i].r<c[j].r)
    				if (dis(c[i].x,c[i].y,c[j].x,c[j].y)<(c[j].r-c[i].r)+e){
    					cov=1;break;}
    		}
    		if(cov)continue;
    		Z[t++]=c[i];
    	}
    
    	k=t;
    	for (i=0;i<k;i++)
    	{
    		for (t=j=0;j<k;j++)
    		{
    			if (i!=j){
    				d=dis(Z[i].x,Z[i].y,Z[j].x,Z[j].y);
    				if(d<(Z[j].r+Z[i].r)){
    					a=(S(d)+S(Z[i].r)-S(Z[j].r))/2/d/Z[i].r;
    					a>1 &&	(a=1);
    					a<-1 && (a=-1);
    					a=acos(a);
    					h=atan2(Z[j].y-Z[i].y,Z[j].x-Z[i].x);
    					p[t++]=h-a;
    					InQue(p,&t);
    					p[t++]=h+a;
    					InQue(p,&t);
    				}
    			}
    		}
    
    		if (t){
    			for (j=0;j<t;j++)
    			{
    				a1=p[j],a2=p[(j==t-1)?0:(j+1)];
    				if(a2<a1)a2+=2*pi;
    				d=(a1+a2)/2;
    				a=Z[i].x+Z[i].r*cos(d),h=Z[i].y+Z[i].r*sin(d);
    				for (q=0;q<k;q++)
    					if(i!=q && dis(a,h,Z[q].x,Z[q].y)<Z[q].r)break;
    				if(q==k){
    					f+=S(Z[i].r)*(a2-a1-sin(a2-a1))/2;
    					f+=((Z[i].x+Z[i].r*cos(a1))*(Z[i].y+Z[i].r*sin(a2))-
    						(Z[i].x+Z[i].r*cos(a2))*(Z[i].y+Z[i].r*sin(a1)))/2;
    				}
    			}
    			
    		}
    		else{
    			f+=pi*S(Z[i].r);}
    	}
    	return f;
    }
    
    int main()
    {
    	int T,R,G,B,i;
    	C c[300],tc[200];
    	double f[7];
    	scanf("%d",&T);
    	for (;T;T--)
    	{
    		memset(f,0,56);
    		scanf("%d %d %d",&R,&G,&B);
    		for (i=0;i<R;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[0]=G_ar(c,R);
    		for (i=R;i<R+G;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[3]=G_ar(c,R+G);
    		for (i=R+G;i<R+G+B;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[6]=G_ar(c,R+G+B);		
    		memcpy(tc,c+R,G*24);
    		f[1]=G_ar(tc,G);
    		memcpy(tc,c+R+G,B*24);
    		f[2]=G_ar(tc,B);
    		memcpy(tc,c+R,(G+B)*24);
    		f[5]=G_ar(tc,G+B);
    		memcpy(c+R,c+(R+G),B*24);
    		f[4]=G_ar(c,R+B);
    		
    		f[3]=f[0]+f[1]-f[3];
    		f[4]=f[0]+f[2]-f[4];
    		f[5]=f[1]+f[2]-f[5];
    		f[6]=f[6]-f[0]-f[1]-f[2]+f[3]+f[4]+f[5];
    		printf("%.2f %.2f %.2f %.2f %.2f %.2f %.2f\n",f[0]-f[3]-f[4]+f[6]+e,
    			f[1]-f[3]-f[5]+f[6]+e,f[2]-f[4]-f[5]+f[6]+e,f[6]+e,f[3]-f[6]+e,f[4]-f[6]+e,f[5]-f[6]+e);
    	}
    }

    优化方案二
    #include <stdio.h>
    #include "string.h"
    #include "math.h"
    
    #define S(x) (x)*(x)
    const double pi=3.141592653589,e=1e-9;
    
    typedef struct{
    	double x,y,r;
    }C;
    
    typedef struct{
    	double b,e;
    }A;
    
    double dis(const double a,const double b,const double c,const double d){
    	return sqrt(S(a-c)+S(b-d));}
    
    void InQue(A p[],int *t,A nt)
    {
    	int i,j;
    	for (i=0;i<*t;i++)
    	{
    		if(nt.b>p[i].e)continue;
    		else if(nt.b>p[i].b){
    			if (nt.e<p[i].e){
    				for (j=*t;j>i+1;j--)
    					p[j]=p[j-1];
    				p[i+1].b=nt.e;
    				p[i+1].e=p[i].e;
    				p[i].e=nt.b;
    				(*t)++;}
    			else{
    				p[i].e=nt.b;}
    		}
    		else{
    			if (nt.e>p[i].e-e){
    				for (j=i;j<*t-1;j++)
    					p[j]=p[j+1];
    				(*t)--,i--;}
    			else if(nt.e>p[i].b){
    				p[i].b=nt.e;}
    		}
    	}
    }
    
    double G_ar(C c[],int k)
    {
    	char cov;
    	int i,j,t=0;
    	C Z[300];
    	A p[10],nt;
    	double f=0,d,h,a;
    	for (i=0;i<k;i++)
    	{
    		if(c[i].r==0)continue;
    		cov=0;
    		for (j=i+1;j<k;j++)
    		{
    			if(c[i].x==c[j].x && c[i].y==c[j].y && c[i].r==c[j].r){
    				cov=1;break;}
    		}
    		if(cov)continue;
    		for (j=0;j<k;j++)
    		{
    			if (i!=j && c[i].r<c[j].r)
    				if (dis(c[i].x,c[i].y,c[j].x,c[j].y)<(c[j].r-c[i].r)+e){
    					cov=1;break;}
    		}
    		if(cov)continue;
    		Z[t++]=c[i];
    	}
    
    	k=t;
    	for (i=0;i<k;i++)
    	{
    		t=1,p[0].b=0,p[0].e=2*pi;
    		for (j=0;j<k;j++)
    		{
    			if (i!=j){
    				d=dis(Z[i].x,Z[i].y,Z[j].x,Z[j].y);
    				if(d<(Z[j].r+Z[i].r)){
    					a=(S(d)+S(Z[i].r)-S(Z[j].r))/2/d/Z[i].r;
    					a>1 && (a=1);
    					a<-1 && (a=-1);
    					a=acos(a);
    					h=atan2(Z[j].y-Z[i].y,Z[j].x-Z[i].x);
    					if (h-a<-e && h+a>e)
    					{
    						nt.b=h-a+2*pi,nt.e=2*pi;
    						InQue(p,&t,nt);
    						nt.b=0,nt.e=h+a;
    						InQue(p,&t,nt);
    					}
    					else
    					{
    						nt.b=h-a,nt.b<-e && (nt.b+=2*pi);
    						nt.e=h+a,nt.e<e && (nt.e+=2*pi);
    						InQue(p,&t,nt);
    					}
    				}
    			}
    		}
    
    		for (j=0;j<t;j++)
    		{
    			a=p[j].b,h=p[j].e;
    			f+=S(Z[i].r)*(h-a-sin(h-a))/2;
    			f+=((Z[i].x+Z[i].r*cos(a))*(Z[i].y+Z[i].r*sin(h))-
    				(Z[i].x+Z[i].r*cos(h))*(Z[i].y+Z[i].r*sin(a)))/2;
    		}
    	}
    	return f;
    }
    
    int main()
    {
    	int T,R,G,B,i;
    	C c[300];
    	double f[7];
    	scanf("%d",&T);
    	for (;T;T--)
    	{
    		memset(f,0,56);
    		scanf("%d %d %d",&R,&G,&B);
    		for (i=0;i<R;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[0]=G_ar(c,R);
    		for (i=R;i<R+G;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[3]=G_ar(c,R+G);
    		for (i=R+G;i<R+G+B;i++)
    			scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
    		f[6]=G_ar(c,R+G+B);
    		f[1]=G_ar(c+R,G);
    		f[2]=G_ar(c+R+G,B);
    		f[5]=G_ar(c+R,G+B);
    		memcpy(c+R,c+(R+G),B*24);
    		f[4]=G_ar(c,R+B);
    		
    		f[3]=f[0]+f[1]-f[3];
    		f[4]=f[0]+f[2]-f[4];
    		f[5]=f[1]+f[2]-f[5];
    		f[6]=f[6]-f[0]-f[1]-f[2]+f[3]+f[4]+f[5];
    		printf("%.2f %.2f %.2f %.2f %.2f %.2f %.2f\n",f[0]-f[3]-f[4]+f[6]+e,
    			f[1]-f[3]-f[5]+f[6]+e,f[2]-f[4]-f[5]+f[6]+e,f[6]+e,f[3]-f[6]+e,f[4]-f[6]+e,f[5]-f[6]+e);
    	}
    }

    调试经验
    优化方案二在f[6]崩溃
    崩溃原因在浮点数等于的处理不对,下面这句话
         else
         {
          nt.b=h-a,nt.b<-e && (nt.b+=2*pi);
          nt.e=h+a,nt.e<-e && (nt.e+=2*pi);
          InQue(p,&t,nt);
         }
    改为下面,表示
         else
         {
          nt.b=h-a,nt.b<-e && (nt.b+=2*pi);
          nt.e=h+a,nt.e<e && (nt.e+=2*pi);
          InQue(p,&t,nt);
         }
    展开全文
  • Multithreading support

    2020-12-04 19:18:18
    <p>We recently switched to a multi-threaded worker model approach and noticed issues from <code>flask_dance</code> (technically from the lazy library): <pre><code> KeyError: 'session' File &#...
  • B,从Blues到HIP-POP、从House Dance到DubStep。 Logic Pro X 是有史以来最高级的 Logic 版本。围绕颇具现代感的界面而开发的全新高级工具可用于专业的音乐创作、编辑以及混音,以快速获得创造性的结果并在需要

    Logic Pro X版是苹果公司设计的一款专业音频制作软件,作为 Mac 上功能完备的专业录音室,Logic Pro X为音乐人提供了从创作第一个音符到完成最后的母带所需的一切。它为您带来的软件乐器与音频处理插件足以让您制作任何风格的音乐!从POP到R&B,从Blues到HIP-POP、从House Dance到DubStep。

    Logic Pro X 是有史以来最高级的 Logic 版本。围绕颇具现代感的界面而开发的全新高级工具可用于专业的音乐创作、编辑以及混音,以快速获得创造性的结果并在需要时助您一臂之力。Logic Pro X 包含种类多样的乐器、效果和循环,形成了一个完整的工具包,可让您创作无比动听的音乐。

    传送门:https://mac.orsoon.com/Mac/156148.html

    功能特色

    1、强大的界面
    借助 Track Stack 来整合并控制多个轨道,或创建种类丰富、层次分明的乐器
    使用 Smart Control 一步便可处理多个插件和参数
    使用增强的混音器来更有效率地移动、拷贝和旁通通道插入
    使用编配标记来快速重新排列乐曲章节的顺序,并尝试新的想法
    自动存储让您的作品安全无忧
    64 位架构支持包含数百个轨道和采样乐器的大型项目
     
    2、专业的音乐创作
    借助 Flex Pitch 来修正不协调的声乐,并更改已录制音频的旋律
    使用 Flex Time 轻松处理任意录音的时序和速度
    录制并无缝设定一个或多个轨道的入点和出点
    使用汇整折叠夹来整理汇整并通过快速扫动伴奏来快速构建伴奏
    不仅限于轨道,自动化也可以是片段的一部分,让您更易于创造性地使用各种效果
    在 iPad 上使用 Logic Remote 在房间的任意位置创建并混音音乐
    借助于内建乐谱编辑器,创作简单的前置表或管弦乐乐谱
    使用一整套编辑器和工具来创建和编辑 MIDI
    3、鼓乐创作
    使用 Drummer(虚拟演奏者及节拍制作人)来创建逼真的原声、电子音乐或嘻哈鼓类轨道
    选择 28 个特色分明的 Drummer 中的任意一个,每个都能按照您的指挥演奏数以百万计的独特音乐轨道
    通过 Drum Kit Designer,您可以使用各种经过深度采样以及专业混音的小军鼓、嗵嗵鼓、脚鼓、竖钹和铙钹来构建自己的原声架子鼓
    使用 Drum Machine Designer 来自定电子乐节拍的声音
    4、键盘与合成器
    一系列提供模拟、波表、调频、加法、粒子、频谱和建模合成的合成器,能极大地激发您的灵感
    通过终极样本处理合成器 Alchemy,快速查找声音或创建独一无二的新声音
    使用 EXS24 采样器来弹奏或创建各种采样丰富的乐器
    通过自动琶音器将简单的和弦立即转换成精彩的演奏
    使用九种 MIDI 插件,简单的想法也可变为精工细作的演奏
    使用 Retro Synth 来创建经典的 70 年代和 80 年代风格的合成器轨道
    使用 Vintage B3、Vintage Electric Piano 和 Vintage Clav 弹奏古典键盘的仿真模型
    5、吉他和贝司设备
    通过 Amp Designer 使用古典与现代放大器、音箱和麦克风来构建自己的吉他或贝司设备
    使用一系列延迟、失真和调制踏脚转盘来设计自定义的踏板
    只需点按一下即可访问调音器来快速合调
    6、创意和作品效果
    使用 Space Designer 回旋混响器来通过逼真的原声空间播放声音
    使用一系列多拍子、老式磁带和立体声延迟效果
    使用各种调制效果给轨道添加运动效果
    7、声音资源库
    超过 1800 个乐器和效果 Patch
    750 多种精密的采样乐器
    4600 个 Apple Loops,分为现代都市和电子类型
    8、打造你自己的音效阵容
    LogicPro 提供了海量插件和音效,让你的创作灵感鲜活呈现。通过 Arpeggiator 添加和弦,尝试新的音效。这里有 1800 种 Patch.乐器.人声音效.低音和声.铜管乐器.弦乐器.木管乐器以及更多音效供你试听。还有 3600 多种电子和都市循环乐段等你去发掘,包括嘻哈.Electro House.Dubstep.现代 R&B.Tech House.Deep House 和 Chillwave,涵盖各种音乐流派。
    9、声音资源库和循环乐段
    声音资源库与循环乐段选集由世界顶尖的创意音效设计师们精心打造。Patch 架构让你能在不同乐器上叠加各种效果,从而获得更精细的成果。而资源库中的一系列 Patch 能够将 Arpeggiator.其他 MIDI 插件和 Track Stack 等 LogicPro 的功能发挥得淋漓尽致。每个 Patch 均搭载一套定制化的 Smart Control,这样你就能轻松创建理想的音效。如果你制作了一款出色的音效,可将它保存为一个 Patch。而你保存的不仅是音轨或多声道的 Track Stack,还包括所有与之相关的处理方式.混音器发送路径和 Smart Control 选项。
    10、塑造音形,瞬间成形
    Smart Control 是开创性的声音塑造工具,让你无需埋头于多个插件界面中,也可以进行简单或精细的创意修改。一项 Smart Control 每次可调整数个插件参数,因此实现复杂的转换变得简单多了。而且你可对 Smart Control 映射进行自定义,从而适应你的工作流程。只需从各种主题的旋钮.按钮及背板中进行选择,即可反映出被控制的乐器类型或效果。
    11、理想的录音方式
    流畅的穿插录音.自动 Take 管理.原始 24-bit/192kHz 音频支持,LogicPro 令其全部易于执行,也便于撤销。250 多个音轨和数百种插件的运行能力,令完成项目所需的一切,总在你的掌握之中。独特的低延迟模式可临时消除插件引起的延迟,让乐手发挥出理想表现。如果搭配 Logic Remote,你还可以在房间任意角落使用 iPad 进行操作。
    12、创建.归整和管理音轨
    将多个相关的音轨 (比如所有的鼓声和人声) 整合到一个被称为 Track Stack 的音轨格式中,从而使你的制作更加有条不紊。选择将你的音轨发送到新的辅助音轨,进行便捷的子混音。也可使用多个 Track Stack 来创建丰富.分层或分离式乐器,以便于管理.保存和重复使用。而且你的 Track Stack 可以折叠起来,从而简化界面,也可随时扩展开来,提供更多控制。
    13、兼容性
    使用与 Audio Units 兼容的第三方插件来扩展您的乐器和效果资源库
    导入并导出 XML 以支持 Final Cut Pro X 工作流程
    导出您的乐曲并直接分享到 SoundCloud
    打开 Logic 5 或更高版本的项目
    展开全文
  • When I then select music, I would expect to see a set of new child categories, such as Rock, Dance, Pop ect. However at the moment I am just presented with all of the individual products. <p>How to...
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    准备 以下のデータベースを作成せよ CREATE DATABASE cd_app ...(2, 'J-POP'), (3, '演歌'), (4, 'DANCE'); 光盘 CREATE TABLE `cds` ( `id` int(11) PRIMARY KEY AUTO_INCREMENT, `title` varchar(255) NOT
  • 2. Stops the disable/re-enable dance when showing a view controller. <p>I don't see any need to disable the back button during a view controller transition. As far as I know, iOS disables ...
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  • Ahu Club/Dance<br> Alex Ferreira Pop/Rock <br> Alex Winston Pop/Rock <br> Ali Azimi Pop/Rock <br> Alphamama Pop/Rock <br> Amaryllis International <br> ... Yomo Toro...
  • ChaeYeon 如何实现呢

    2020-01-16 12:01:54
    Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
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  • Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
  • Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
  • Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
  • ChaeYeon

    2017-10-31 12:34:16
    Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
  • Post-disco, Rhythm and Blues, Funk, Dance-pop', duration: this.whoosh.getDuration(), description: 'Billie Jean is a song by American singer Michael Jackson. It is the second single from the ...
  • [3, 4, 6]</code> to detect from a mixed bunch over many genres (but mostly pop / rock / soul / dance music). <p>The vast majority is in common (4) time, with a few 6/8 plus a few "difficult" ...
  • Emails application

    2020-12-01 11:43:03
    - If you use multiple mails accounts, you have to wait for all your accounts to be checked before clicking on a message - The TCP code I used allow only one socket, so I need to dance around to ...
  • Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in ...
  • Post-disco, Rhythm and Blues, Funk, Dance-pop', duration: this.whoosh.getDuration(), }) MusicControl.enableControl('play', false) MusicControl.enableControl('pause', true) ...
  • slow queries

    2020-11-27 07:58:45
    DANCE','DANCE`','DISCO','DJ_TOOLS','ELECTRONIC','ELETRONIC','FRENCH','FUNKnSOUL','GERMAN','HIPHOPnRAP','HOUSE',&...
  • Post-disco, Rhythm and Blues, Funk, Dance-pop', duration: 294, // (Seconds) description: '', // Android Only color: 0xFFFFFF, // Notification Color - Android Only date: '1983-01-02...

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