HDU 5130 Signal Interference(计算几何 + 模板)
HDU 5130 Signal Interference(计算几何 + 模板)
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5130
Description
Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.
Input
There are no more than 100 test cases in the input.
In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point.
Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order.
The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.
Output
For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'.
Your solution will be accepted if its absolute error or relative error is no more than 10-6.
This problem is special judged.
Sample Input
4 0.5000
-1 -1
1 -1
1 1
-1 1
0 0
-1 0
Sample Output
Case 1: 0.2729710441
题意：
给你n个点按照顺时针或者逆时针排序围成多边形，A,B点，让你计算从某点到B点的距离是到A距离的K倍，求这个图形和多边形的相交的面积。
题解：
求的点带入，化简就是一个圆，然后就是圆和多边形的面积交。套模板。
代码：
#include <bits/stdc++.h>
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=0, double y=0):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c,double r):c(c),r(r) {}
Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }
bool OnSegment(Point P, Point A, Point B) {
return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
}
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
double DistanceToLine(Point P, Point A, Point B){
Vector v1 = B-A, v2 = P-A;
return fabs(Cross(v1,v2)) / Length(v1);
}
Point DisP(Point A, Point B){
return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
return max(A.x,B.x) >= min(C.x,D.x) &&
max(C.x,D.x) >= min(A.x,B.x) &&
max(A.y,B.y) >= min(C.y,D.y) &&
max(C.y,D.y) >= min(A.y,B.y) &&
dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}
Point Zero = Point(0,0);
//sum_ans !!!!!!!fabs()
double TriAngleCircleInsection(Circle C, Point A, Point B)
{
Vector OA = A-C.c, OB = B-C.c;
Vector BA = A-B, BC = C.c-B;
Vector AB = B-A, AC = C.c-A;
double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
if(dcmp(Cross(OA,OB)) == 0) return 0;
if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
else if(DOB < r && DOA >= r) {
double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
}
else if(DOB >= r && DOA < r) {
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
}
else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
if(Dot(OA,OB) < 0) {
if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
else return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
}
else return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
}
else {
double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
}
}
Point s[600],A,B ;
int main()
{
int n ;
int _t = 0;
while (~scanf("%d",&n)){
double k ;
_t++ ;
scanf("%lf",&k) ;
for (int i = 1;i <= n; i++)
s[i].input();
A.input();B.input();
s[n+1] = s[1];
double D,E,F;
D = (2.0*k*k*A.x - 2.0*B.x)/(1.0-k*k) ;
E = (2.0*k*k*A.y - 2.0*B.y)/(1.0-k*k) ;
F = (B.x*B.x+B.y*B.y-k*k*(A.x*A.x+A.y*A.y))/(1.0-k*k) ;
Circle C = Circle(Point(D*(-0.5),E*(-0.5)),sqrt(D*D+E*E-4.0*F)*0.5) ;
double ans = 0.0;
for (int i = 1; i <= n; i++){
ans = ans + TriAngleCircleInsection(C,s[i],s[i+1]) ;
}
printf("Case %d: %.10lf\n",_t,fabs(ans)) ;
}
return 0;
}
HDU 5130 Signal Interference(计算几何 + 模板)的更多相关文章
- HDU 5130 Signal Interference --计算几何,多边形与圆的交面积
题意: 求所有满足PB <= k*PA 的P所在区域与多边形的交面积. 解法: 2014广州赛区的银牌题,当时竟然没发现是圆,然后就没做出来,然后就gg了. 圆的一般式方程: 设A(x1,y1) ...
- hdu 4667 Building Fence < 计算几何模板>
//大白p263 #include <cmath> #include <cstdio> #include <cstring> #include <string ...
- LA 7072 Signal Interference 计算几何 圆与多边形的交
题意: 给出一个\(n\)个点的简单多边形,和两个点\(A, B\)还有一个常数\(k(0.2 \leq k < 0.8)\). 点\(P\)满足\(\left | PB \right | \l ...
- HDU5130 Signal Interference
/* HDU5130 Signal Interference http://acm.hdu.edu.cn/showproblem.php?pid=5130 计算几何 圆与多边形面积交 * */ #in ...
- lrj计算几何模板
整理了一下大白书上的计算几何模板. #include <cstdio> #include <algorithm> #include <cmath> #include ...
- UVA 12304 - 2D Geometry 110 in 1! - [平面几何基础题大集合][计算几何模板]
题目链接:https://cn.vjudge.net/problem/UVA-12304 题意: 作为题目大合集,有以下一些要求: ①给出三角形三个点,求三角形外接圆,求外接圆的圆心和半径. ②给出三 ...
- HDU - 5130 ：Signal Interference （多边形与圆的交）
pro:A的监视区域是一个多边形. 如果A的监视区的内满足到A的距离到不超过到B的距离的K倍的面积大小.K<1 sol:高中几何体经验告诉我们满足题意的区域是个圆,那么就是求圆与多边形的交. # ...
- hdu 5130(2014广州 圆与多边形相交模板)
题意:一个很多个点p构成的多边形,pb <= pa * k时p所占区域与多边形相交面积 设p(x,y), (x - xb)^2+(y - yb)^2 / (x - xa)^2+(y ...
- hdu 3060 Area2 (计算几何模板)
Problem Description 小白最近又被空军特招为飞行员,参与一项实战演习.演习的内容还是轰炸某个岛屿(这次的岛屿很大,很大很大很大,大到炸弹怎么扔都能完全在岛屿上引爆),看来小白确实是飞 ...
随机推荐
- 飘逸的python - zlib压缩存到数据库
当每天有大量的数据存到kv数据库中去,且value数据很大,于是想压缩后再存进去. 之前提到了 gzip压缩,为什么不直接用gzip呢. 其实更确切的说gzip是一种文件格式,它压缩成gzip文件,而 ...
- STM8S ADC初始化设置及应用
//ADC通道号定义 #define ADC_Chanel0 (unsigned char)0x00 #define ADC_Chanel1 (unsigned char)0x01 #define A ...
- C++中const用法详解
本文主要内容来自CSDN论坛: http://bbs.csdn.net/topics/310007610 我做了下面几点补充. 补充: 1. 用const声明全局变量时, 该变量仅在本文件内可见, 类 ...
- nginx+apache+mysql+php+memcache+squid搭建集群web环境
服务器的大用户量的承载方案 一.前言 二.编译安装 三. 安装MySQL.memcache 四. 安装Apache.PHP.eAccelerator.php-memcache 五. 安装Squid 六 ...
- dom4解析xml格式文件实例
以下给4种常见的xml文件的解析方式的分析对比: DOM DOM4J JDOM SAX 解析XML文件的几种方式和区别答: Dom解析 在内存中创建一个DOM树,该结构通常需要加载整个文档然后才能做工 ...
- c语言中细节注意（初级）
/* 编写如下函数,不使用下标运算符,返回字符串str中字符c的个数 (若不存在则为0). */ #include <stdio.h> int str_chnum(const char * ...
- jquery中这句 .stop(false,true); 什么意思
.stop 是jQuery中用于控制页面动画效果的方法.运行之后立刻结束当前页面上的动画效果.stop在新版jQuery中添加了2个参数:第一个参数的意思是是否清空动画序列,也就是stop的是当前元素 ...
- Css3新特性应用之过渡与动画
目录 背景与边框第一部分 背景与边框第二部分 形状 视觉效果 字体排印 用户体验 结构与布局 过渡与动画 源码下载 一.缓动效果 学习和利用贝塞尔曲线,默认支持ease,ease-in,ease-ou ...
- springboot学习笔记-1 第一个springboot示例
springboot是一个微框架,其设计的目的是为了简化spring框架的搭建和配置过程.从而使开发人员不再需要定义样板化的配置.下面是springboot的入门案例:它演示了利用springboot ...
- 【LeetCode】31. Next Permutation
Implement next permutation, which rearranges numbers into the lexicographically next greater permuta ...
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5130
Description
Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.
Input
There are no more than 100 test cases in the input.
In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point.
Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order.
The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.
Output
For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'.
Your solution will be accepted if its absolute error or relative error is no more than 10-6.
This problem is special judged.
Sample Input
4 0.5000
-1 -1
1 -1
1 1
-1 1
0 0
-1 0
Sample Output
Case 1: 0.2729710441
题意：
给你n个点按照顺时针或者逆时针排序围成多边形，A,B点，让你计算从某点到B点的距离是到A距离的K倍，求这个图形和多边形的相交的面积。
题解：
求的点带入，化简就是一个圆，然后就是圆和多边形的面积交。套模板。
代码：
#include <bits/stdc++.h>
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=0, double y=0):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c,double r):c(c),r(r) {}
Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }
bool OnSegment(Point P, Point A, Point B) {
return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
}
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
double DistanceToLine(Point P, Point A, Point B){
Vector v1 = B-A, v2 = P-A;
return fabs(Cross(v1,v2)) / Length(v1);
}
Point DisP(Point A, Point B){
return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
return max(A.x,B.x) >= min(C.x,D.x) &&
max(C.x,D.x) >= min(A.x,B.x) &&
max(A.y,B.y) >= min(C.y,D.y) &&
max(C.y,D.y) >= min(A.y,B.y) &&
dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}
Point Zero = Point(0,0);
//sum_ans !!!!!!!fabs()
double TriAngleCircleInsection(Circle C, Point A, Point B)
{
Vector OA = A-C.c, OB = B-C.c;
Vector BA = A-B, BC = C.c-B;
Vector AB = B-A, AC = C.c-A;
double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
if(dcmp(Cross(OA,OB)) == 0) return 0;
if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
else if(DOB < r && DOA >= r) {
double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
}
else if(DOB >= r && DOA < r) {
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
}
else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
if(Dot(OA,OB) < 0) {
if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
else return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
}
else return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
}
else {
double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
}
}
Point s[600],A,B ;
int main()
{
int n ;
int _t = 0;
while (~scanf("%d",&n)){
double k ;
_t++ ;
scanf("%lf",&k) ;
for (int i = 1;i <= n; i++)
s[i].input();
A.input();B.input();
s[n+1] = s[1];
double D,E,F;
D = (2.0*k*k*A.x - 2.0*B.x)/(1.0-k*k) ;
E = (2.0*k*k*A.y - 2.0*B.y)/(1.0-k*k) ;
F = (B.x*B.x+B.y*B.y-k*k*(A.x*A.x+A.y*A.y))/(1.0-k*k) ;
Circle C = Circle(Point(D*(-0.5),E*(-0.5)),sqrt(D*D+E*E-4.0*F)*0.5) ;
double ans = 0.0;
for (int i = 1; i <= n; i++){
ans = ans + TriAngleCircleInsection(C,s[i],s[i+1]) ;
}
printf("Case %d: %.10lf\n",_t,fabs(ans)) ;
}
return 0;
}
题意: 求所有满足PB <= k*PA 的P所在区域与多边形的交面积. 解法: 2014广州赛区的银牌题,当时竟然没发现是圆,然后就没做出来,然后就gg了. 圆的一般式方程: 设A(x1,y1) ...
//大白p263 #include <cmath> #include <cstdio> #include <cstring> #include <string ...
题意: 给出一个\(n\)个点的简单多边形,和两个点\(A, B\)还有一个常数\(k(0.2 \leq k < 0.8)\). 点\(P\)满足\(\left | PB \right | \l ...
/* HDU5130 Signal Interference http://acm.hdu.edu.cn/showproblem.php?pid=5130 计算几何 圆与多边形面积交 * */ #in ...
整理了一下大白书上的计算几何模板. #include <cstdio> #include <algorithm> #include <cmath> #include ...
题目链接:https://cn.vjudge.net/problem/UVA-12304 题意: 作为题目大合集,有以下一些要求: ①给出三角形三个点,求三角形外接圆,求外接圆的圆心和半径. ②给出三 ...
pro:A的监视区域是一个多边形. 如果A的监视区的内满足到A的距离到不超过到B的距离的K倍的面积大小.K<1 sol:高中几何体经验告诉我们满足题意的区域是个圆,那么就是求圆与多边形的交. # ...
题意:一个很多个点p构成的多边形,pb <= pa * k时p所占区域与多边形相交面积 设p(x,y), (x - xb)^2+(y - yb)^2 / (x - xa)^2+(y ...
Problem Description 小白最近又被空军特招为飞行员,参与一项实战演习.演习的内容还是轰炸某个岛屿(这次的岛屿很大,很大很大很大,大到炸弹怎么扔都能完全在岛屿上引爆),看来小白确实是飞 ...
当每天有大量的数据存到kv数据库中去,且value数据很大,于是想压缩后再存进去. 之前提到了 gzip压缩,为什么不直接用gzip呢. 其实更确切的说gzip是一种文件格式,它压缩成gzip文件,而 ...
//ADC通道号定义 #define ADC_Chanel0 (unsigned char)0x00 #define ADC_Chanel1 (unsigned char)0x01 #define A ...
本文主要内容来自CSDN论坛: http://bbs.csdn.net/topics/310007610 我做了下面几点补充. 补充: 1. 用const声明全局变量时, 该变量仅在本文件内可见, 类 ...
服务器的大用户量的承载方案 一.前言 二.编译安装 三. 安装MySQL.memcache 四. 安装Apache.PHP.eAccelerator.php-memcache 五. 安装Squid 六 ...
以下给4种常见的xml文件的解析方式的分析对比: DOM DOM4J JDOM SAX 解析XML文件的几种方式和区别答: Dom解析 在内存中创建一个DOM树,该结构通常需要加载整个文档然后才能做工 ...
/* 编写如下函数,不使用下标运算符,返回字符串str中字符c的个数 (若不存在则为0). */ #include <stdio.h> int str_chnum(const char * ...
.stop 是jQuery中用于控制页面动画效果的方法.运行之后立刻结束当前页面上的动画效果.stop在新版jQuery中添加了2个参数:第一个参数的意思是是否清空动画序列,也就是stop的是当前元素 ...
目录 背景与边框第一部分 背景与边框第二部分 形状 视觉效果 字体排印 用户体验 结构与布局 过渡与动画 源码下载 一.缓动效果 学习和利用贝塞尔曲线,默认支持ease,ease-in,ease-ou ...
springboot是一个微框架,其设计的目的是为了简化spring框架的搭建和配置过程.从而使开发人员不再需要定义样板化的配置.下面是springboot的入门案例:它演示了利用springboot ...
Implement next permutation, which rearranges numbers into the lexicographically next greater permuta ...