Time Limit: 2000MS | Memory Limit: 65536K | |||
Total Submissions: 7687 | Accepted: 2075 | Special Judge |
Description
Your task is to write a part of this software that determines the position of the end of the n-th segment after each command. The state of the crane is determined by the angles between consecutive segments. Initially, all of the angles are straight, i.e., 180o. The operator issues commands that change the angle in exactly one joint.
Input
The first line of each instance consists of two integers 1 ≤ n ≤10 000 and c 0 separated by a single space -- the number of segments of the crane and the number of commands. The second line consists of n integers l1,..., ln (1 li 100) separated by single spaces. The length of the i-th segment of the crane is li. The following c lines specify the commands of the operator. Each line describing the command consists of two integers s and a (1 ≤ s < n, 0 ≤ a ≤ 359) separated by a single space -- the order to change the angle between the s-th and the s + 1-th segment to a degrees (the angle is measured counterclockwise from the s-th to the s + 1-th segment).
Output
The outputs for each two consecutive instances must be separated by a single empty line.
Sample Input
2 1 10 5 1 90 3 2 5 5 5 1 270 2 90
Sample Output
5.00 10.00 -10.00 5.00 -5.00 10.00
Source
vy[i]=vy[chl]+ (sin(angi)*vx[chr]+cos(angi)*vy[chr]);
1 #include <iostream> 2 #include <cmath> 3 #include <cstdio> 4 #define MAX_N 10000 5 #define M_PI 3.14159265358979323846 6 using namespace std; 7 8 const int ST_SIZE=(1<<15)-1; 9 //输入 10 int N,C; 11 int L[MAX_N+5]; 12 int S[MAX_N+5],A[MAX_N+5]; 13 14 //线段树所维护的数据 15 double vx[ST_SIZE],vy[ST_SIZE]; 16 double ang[ST_SIZE]; 17 18 //为了查询角度的变化而保存的当前角度的数组 19 double prv[MAX_N]; 20 21 //初始化线段树 22 //k是节点的编号,l,r表示当前节点对应的是[l,r]区间 23 void init(int k,int l,int r) 24 { 25 ang[k]=vx[k]=0.0; 26 if(r-l==1) 27 { 28 //叶子节点 29 vy[k]=L[l]; 30 } 31 else 32 { 33 //非叶子节点 34 int chl=k*2+1,chr=k*2+2; 35 init(chl,l,(l+r)/2); 36 init(chr,(l+r)/2,r); 37 vy[k]=vy[chl]+vy[chr]; 38 } 39 } 40 //把s和s+1的角度变为a 41 //v是节点的编号,,l,r表示当前节点对应的是[l,r]区间 42 void change(int s,double a,int v,int l,int r) 43 { 44 if(s<=l) 45 return; 46 else if(s<r) 47 { 48 int chl=v*2+1,chr=v*2+2; 49 int m=(l+r)/2; 50 change(s,a,chl,l,m); 51 change(s,a,chr,m,r); 52 if(s<=m) 53 ang[v]+=a; 54 double s=sin(ang[v]),c=cos(ang[v]); //围绕原点的旋转: 55 vx[v]=vx[chl]+(c*vx[chr]-s*vy[chr]); //x' = x * cos(a) - y * sin(a) 56 vy[v]=vy[chl]+(s*vx[chr]+c*vy[chr]); //y' = x * sin(a) + y * cos(a) 57 } 58 59 } 60 void solve() 61 { 62 //初始化 63 init(0,0,N); 64 for(int i=1;i<N;i++) 65 prv[i]=M_PI; 66 //处理操作 67 for(int i=0;i<C;i++) 68 { 69 int s=S[i]; 70 double a=A[i]/360.0*2*M_PI; //把角度换算为弧度 71 change(s,a-prv[s],0,0,N); 72 prv[s]=a; 73 printf("%.2f %.2f\n",vx[0],vy[0]); 74 75 } 76 77 78 } 79 int main() 80 { 81 while(cin>>N>>C) 82 { 83 for(int i=0;i<N;i++) 84 scanf("%d",&L[i]); 85 for(int i=0;i<C;i++) 86 scanf("%d%d",&S[i],&A[i]); 87 solve(); 88 89 } 90 91 return 0; 92 }
Time Limit: 2000MS | Memory Limit: 65536K | |||
Total Submissions: 7687 | Accepted: 2075 | Special Judge |
Description
Your task is to write a part of this software that determines the position of the end of the n-th segment after each command. The state of the crane is determined by the angles between consecutive segments. Initially, all of the angles are straight, i.e., 180o. The operator issues commands that change the angle in exactly one joint.
Input
The first line of each instance consists of two integers 1 ≤ n ≤10 000 and c 0 separated by a single space -- the number of segments of the crane and the number of commands. The second line consists of n integers l1,..., ln (1 li 100) separated by single spaces. The length of the i-th segment of the crane is li. The following c lines specify the commands of the operator. Each line describing the command consists of two integers s and a (1 ≤ s < n, 0 ≤ a ≤ 359) separated by a single space -- the order to change the angle between the s-th and the s + 1-th segment to a degrees (the angle is measured counterclockwise from the s-th to the s + 1-th segment).
Output
The outputs for each two consecutive instances must be separated by a single empty line.
Sample Input
2 1 10 5 1 90 3 2 5 5 5 1 270 2 90
Sample Output
5.00 10.00 -10.00 5.00 -5.00 10.00
Source
vy[i]=vy[chl]+ (sin(angi)*vx[chr]+cos(angi)*vy[chr]);
1 #include <iostream> 2 #include <cmath> 3 #include <cstdio> 4 #define MAX_N 10000 5 #define M_PI 3.14159265358979323846 6 using namespace std; 7 8 const int ST_SIZE=(1<<15)-1; 9 //输入 10 int N,C; 11 int L[MAX_N+5]; 12 int S[MAX_N+5],A[MAX_N+5]; 13 14 //线段树所维护的数据 15 double vx[ST_SIZE],vy[ST_SIZE]; 16 double ang[ST_SIZE]; 17 18 //为了查询角度的变化而保存的当前角度的数组 19 double prv[MAX_N]; 20 21 //初始化线段树 22 //k是节点的编号,l,r表示当前节点对应的是[l,r]区间 23 void init(int k,int l,int r) 24 { 25 ang[k]=vx[k]=0.0; 26 if(r-l==1) 27 { 28 //叶子节点 29 vy[k]=L[l]; 30 } 31 else 32 { 33 //非叶子节点 34 int chl=k*2+1,chr=k*2+2; 35 init(chl,l,(l+r)/2); 36 init(chr,(l+r)/2,r); 37 vy[k]=vy[chl]+vy[chr]; 38 } 39 } 40 //把s和s+1的角度变为a 41 //v是节点的编号,,l,r表示当前节点对应的是[l,r]区间 42 void change(int s,double a,int v,int l,int r) 43 { 44 if(s<=l) 45 return; 46 else if(s<r) 47 { 48 int chl=v*2+1,chr=v*2+2; 49 int m=(l+r)/2; 50 change(s,a,chl,l,m); 51 change(s,a,chr,m,r); 52 if(s<=m) 53 ang[v]+=a; 54 double s=sin(ang[v]),c=cos(ang[v]); //围绕原点的旋转: 55 vx[v]=vx[chl]+(c*vx[chr]-s*vy[chr]); //x' = x * cos(a) - y * sin(a) 56 vy[v]=vy[chl]+(s*vx[chr]+c*vy[chr]); //y' = x * sin(a) + y * cos(a) 57 } 58 59 } 60 void solve() 61 { 62 //初始化 63 init(0,0,N); 64 for(int i=1;i<N;i++) 65 prv[i]=M_PI; 66 //处理操作 67 for(int i=0;i<C;i++) 68 { 69 int s=S[i]; 70 double a=A[i]/360.0*2*M_PI; //把角度换算为弧度 71 change(s,a-prv[s],0,0,N); 72 prv[s]=a; 73 printf("%.2f %.2f\n",vx[0],vy[0]); 74 75 } 76 77 78 } 79 int main() 80 { 81 while(cin>>N>>C) 82 { 83 for(int i=0;i<N;i++) 84 scanf("%d",&L[i]); 85 for(int i=0;i<C;i++) 86 scanf("%d%d",&S[i],&A[i]); 87 solve(); 88 89 } 90 91 return 0; 92 }