The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains an integer n denoting the number of the rotations. Then n lines follows, each containing 3 real numbers x, y and p, which means rotating around point (x, y) counter-clockwisely by a radian of p.
We promise that the sum of all p‘s is differed at least 0.1 from the nearest multiplier of 2π.
T<=100. 1<=n<=10. 0<=x, y<=100. 0<=p<=2π.
For each test case, print 3 real numbers x, y, p, indicating that the overall rotation is around (x, y) counter-clockwisely by a radian of p. Note that you should print p where 0<=p<2π.
Your answer will be considered correct if and only if for x, y and p, the absolute error is no larger than 1e-5.
1.8088715944 0.1911284056 3.0000000000
对于每一组测试数据，给出一系列旋转命令，每次将整个平面空间绕着某一点逆时针旋转某一个弧度。而所有这些操作的最终结果可以等效为绕着A点旋转P弧度，题目的要求即求出这个坐标点A和弧度P。
当时看到这题解析几何，第一反应就是不想去管它了，因为手头没有趁手的模板可用，本来就数学不好，算起来太过耗时，还容易错，无奈水平有限，能做的题目不多，还是回来写这个了。
既然题目确定了所有这些操作都可以等效为一个，那我们也不需呀去管这个结论是怎么证明出来的了。基本的思想是用一条线段去指代这个平面，每次将线段的两端点都进行一次旋转操作，就相当于整条线段都旋转了，更进一步地就代表平面旋转了。之后分析原始坐标与结果坐标之间的关系即可得出等效结果。
最开始一直WA是忽略了一个问题就是P的范围是0~2Pi，三角形余弦定理计算出的结果肯定是小于Pi的，就是最后必须判断一下旋转角度是否超过了Pi，超了则最终的输出结果应该是2Pi-P。这里要用到判断两点是否在一条直线的同一边。
1 /*
2 ID: Chen Fan
3 PROG: B1002
4 LANG: G++
5 */
6
7 #include<iostream>
8 #include<cstdio>
9 #include<cmath>
10
11 #define Pi 3.141592657
12
13 typedef struct poi
14 {
15 double x,y;
16 } point;
17
18 using namespace std;
19
20 point rotate(point v,point p,double angle)
21 {
22 point ret=p;
23 v.x-=p.x,v.y-=p.y;
24 p.x=cos(angle);
25 p.y=sin(angle);
26 ret.x+=v.x*p.x-v.y*p.y;
27 ret.y+=v.x*p.y+v.y*p.x;
28 return ret;
29 }
30
31 int main()
32 {
33 int t;
34 scanf("%d",&t);
35 for (int tt=1;tt<=t;tt++)
36 {
37 int n;
38 scanf("%d",&n);
39 point p1={20.123,6.678},p2={3.414,10.123};
40 point p0={(p1.x+p2.x)/2,(p1.y+p2.y)/2};
41 for (int i=1;i<=n;i++)
42 {
43 point now;
44 double p;
45 scanf("%lf%lf%lf",&now.x,&now.y,&p);
46 p1=rotate(p1,now,p);
47 p2=rotate(p2,now,p);
48 }
49
50 point p10;
51 p10.x=(p1.x+20.123)/2;
52 p10.y=(p1.y+6.678)/2;
53 double k1=(p1.y-p10.y)/(p1.x-p10.x);
54 k1=-1/k1;
55 double b1=p10.y-k1*p10.x;
56
57 point p20;
58 p20.x=(p2.x+3.414)/2;
59 p20.y=(p2.y+10.123)/2;
60 double k2=(p2.y-p20.y)/(p2.x-p20.x);
61 k2=-1/k2;
62 double b2=p20.y-k2*p20.x;
63
64 double xx=(b2-b1)/(k1-k2);
65 double yy=k1*xx+b1;
66
67 double bb=(xx-3.414)*(xx-3.414)+(yy-10.123)*(yy-10.123);
68 double cc=(xx-p2.x)*(xx-p2.x)+(yy-p2.y)*(yy-p2.y);
69 double aa=(p2.x-3.414)*(p2.x-3.414)+(p2.y-10.123)*(p2.y-10.123);
70 double ct=(bb+cc-aa)/(2*sqrt(bb)*sqrt(cc));
71
72 point p3={2*xx-p0.x,2*yy-p0.y};
73 point p4={(p1.x+p2.x)/2,(p1.y+p2.y)/2};
74
75 double k3=(p3.y-p0.y)/(p3.x-p0.x);
76 double b3=p3.y-k3*p3.x;
77
78 point pp={xx,yy};
79 point p5=rotate(p0,pp,1);
80
81 if ((p4.x*k3+b3-p4.y)*(p5.x*k3+b3-p5.y)>0) printf("%.10lf %.10lf %.10lf\n",xx,yy,acos(ct));
82 else printf("%.10lf %.10lf %.10lf\n",xx,yy,2*Pi-acos(ct));
83 }
84
85 return 0;
86 }
View Code
【E】
Walk
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
【Problem Description】
I used to think I could be anything, but now I know that I couldn‘t do anything. So I started traveling.
The nation looks like a connected bidirectional graph, and I am randomly walking on it. It means when I am at node i, I will travel to an adjacent node with the same probability in the next step. I will pick up the start node randomly (each node in the graph has the same probability.), and travel for d steps, noting that I may go through some nodes multiple times.
If I miss some sights at a node, it will make me unhappy. So I wonder for each node, what is the probability that my path doesn‘t contain it.
【Input】
The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains 3 integers n, m and d, denoting the number of vertices, the number of edges and the number of steps respectively. Then m lines follows, each containing two integers a and b, denoting there is an edge between node a and node b.
T<=20, n<=50, n-1<=m<=n*(n-1)/2, 1<=d<=10000. There is no self-loops or multiple edges in the graph, and the graph is connected. The nodes are indexed from 1.
【Output】
For each test cases, output n lines, the i-th line containing the desired probability for the i-th node.
Your answer will be accepted if its absolute error doesn‘t exceed 1e-5.
【Sample Input】
2
5 10 100
1 2
2 3
3 4
4 5
1 5
2 4
3 5
2 5
1 4
1 3
10 10 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
4 9
【Sample Output】
0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.6993317967
0.5864284952
0.4440860821
0.2275896991
0.4294074591
0.4851048742
0.4896018842
0.4525044250
0.3406567483
0.6421630037
【题意】
随机随机随机...在一张无向图中，随机选定一个起点出发，之后的每一步都是等概率的。有一些点不能被到达，这里要求的就是每一个点不能被到达的概率。
【分析】
概率DP的题目做得太少了
【G】
Osu!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Special Judge
【Problem Description】
Osu! is a famous music game that attracts a lot of people. In osu!, there is a performance scoring system, which evaluates your performance. Each song you have played will have a score. And the system will sort all you scores in descending order. After that, the i-th song scored ai will add 0.95^(i-1)*ai to your total score.
Now you are given the task to write a calculator for this system.
【Input】
The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains an integer n, denoting the number of songs you have played. The second line contains n integers a1, a2, ..., an separated by a single space, denoting the score of each song.
T<=20, n<=50, 1<=ai<=500.
【Output】
For each test case, output one line for the answer.
Your answers will be considered correct if its absolute error is smaller than 1e-5.
【Sample Input】
【Sample Output】
【分析】
水过了....
