SRM 401(1-250pt, 1-500pt)

DIV1 250pt

题意：给一个整数f，则这样的正整数整数数列称为好数列：数列元素a0 >= a1 >= a2...，且a0<= f, a1 <= f-1, a2 <= f-2....。求给定f后，好数列有多少个。

解法：首先对于f = 3，{2, 1}为好数列，则补0称为{2, 1, 0}。然后，所有好数列长度都变为f。直接dp就好，d[i][j]表示长度为i，第一个元素(最大的)为j的数列有多少个。

　　　d[i][j] = sum(d[i-1][k])其中k <= j。

Ps:官方题解写着还可以用Catalan数做，只不过我想不通为什么可以...

tag:DP

 1 // BEGIN CUT HERE
 2 /*
 3  * Author:  plum rain
 4  * score :
 5  */
 6 /*
 7 
 8  */
 9 // END CUT HERE
10 #line 11 "FIELDDiagrams.cpp"
11 #include <sstream>
12 #include <stdexcept>
13 #include <functional>
14 #include <iomanip>
15 #include <numeric>
16 #include <fstream>
17 #include <cctype>
18 #include <iostream>
19 #include <cstdio>
20 #include <vector>
21 #include <cstring>
22 #include <cmath>
23 #include <algorithm>
24 #include <cstdlib>
25 #include <set>
26 #include <queue>
27 #include <bitset>
28 #include <list>
29 #include <string>
30 #include <utility>
31 #include <map>
32 #include <ctime>
33 #include <stack>
34 
35 using namespace std;
36 
37 #define CLR(x) memset(x, 0, sizeof(x))
38 #define CLR1(x) memset(x, -1, sizeof(x))
39 #define PB push_back
40 #define SZ(v) ((int)(v).size())
41 #define zero(x) (((x)>0?(x):-(x))<eps)
42 #define out(x) cout<<#x<<":"<<(x)<<endl
43 #define tst(a) cout<<#a<<endl
44 #define CINBEQUICKER std::ios::sync_with_stdio(false)
45 
46 typedef vector<int> VI;
47 typedef vector<string> VS;
48 typedef vector<double> VD;
49 typedef pair<int, int> pii;
50 typedef long long int64;
51 
52 const double eps = 1e-8;
53 const double PI = atan(1.0)*4;
54 const int maxint = 2139062143;
55 
56 int64 d[50][50];
57 
58 class FIELDDiagrams
59 {
60     public:
61         long long countDiagrams(int f){
62             ++ f;
63             for (int i = 0; i < f; ++ i)
64                 d[1][i] = 1;
65             for (int i = 2; i < f; ++ i){
66                 for (int j = 0; j <= f-i; ++ j){
67                     d[i][j] = 0;
68                     for (int k = j; k <= f-i+1; ++ k)
69                         d[i][j] += d[i-1][k];
70                 }
71             }
72             return d[f-1][0] + d[f-1][1] - 1;
73         }
74         
75 // BEGIN CUT HERE
76     public:
77     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); }
78     private:
79     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
80     void verify_case(int Case, const long long &Expected, const long long &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
81     void test_case_0() { int Arg0 = 2; long long Arg1 = 4LL; verify_case(0, Arg1, countDiagrams(Arg0)); }
82     void test_case_1() { int Arg0 = 3; long long Arg1 = 13LL; verify_case(1, Arg1, countDiagrams(Arg0)); }
83     void test_case_2() { int Arg0 = 5; long long Arg1 = 131LL; verify_case(2, Arg1, countDiagrams(Arg0)); }
84 
85 // END CUT HERE
86 
87 };
88 
89 // BEGIN CUT HERE
90 int main()
91 {
92 //    freopen( "a.out" , "w" , stdout );    
93     FIELDDiagrams ___test;
94     ___test.run_test(-1);
95        return 0;
96 }
97 // END CUT HERE
View Code

 

DIV2 550pt

题意：有两个物体运动轨迹一个为x1 = cos(t1), y1 = sin(t1), z1 = t1，另一个物体运动轨迹为x2 = vx*t + x0, y2 = vy*t + y0, z2 = vz*t + z0。只要两个运动轨迹有交点即可，不需要t1 = t2。重合交点算一个。如果有多个交点输出{0, 0, 0}，一个交点输出交点坐标，没有交点输出空集。

解法：理解题意理解了半天，还搞错了。。。上网看了下才知道。。。然后，代码本身不太好写，写之前想得太简单又写错了，于是这道题做了好久- -.........

　　　首先，发现x1与y1的关系为x1*x1 + y1*y1 = 1，则x2*x2 + y2*y2 = 1，这样就变成了一元二次方程，当然，解出来的根还不一定满足题意，还要再判断。

Ps:官方题解说还有一种方法，就是有一个结论，要相交必然z为1/2的倍数。至于为什么，题解没有说，给了论坛的地址，看见里面的人在用英语讨论，也并不简单，我就放弃了。

tag:math

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "ParticleCollision.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define CLR(x) memset(x, 0, sizeof(x))
 38 #define CLR1(x) memset(x, -1, sizeof(x))
 39 #define PB push_back
 40 #define SZ(v) ((int)(v).size())
 41 #define zero(x) (((x)>0?(x):-(x))<eps)
 42 #define out(x) cout<<#x<<":"<<(x)<<endl
 43 #define tst(a) cout<<#a<<endl
 44 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 45 
 46 typedef vector<int> VI;
 47 typedef vector<string> VS;
 48 typedef vector<double> VD;
 49 typedef pair<int, int> pii;
 50 typedef long long int64;
 51 
 52 const double eps = 1e-11;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 
 56 bool ok(double t, int A, int B, int C, int a, int b, int c)
 57 {
 58     double x2 = A*t + a, y2 = B*t + b, z = C*t + c;
 59     double x1 = cos(z*atan(1.0)*4), y1 = sin(z*atan(1.0)*4);
 60     if (fabs(x1-x2) < eps && fabs(y1-y2) < eps) return 1;
 61     return 0;
 62 }
 63 
 64 class ParticleCollision
 65 {
 66     public:
 67         vector <double> collision(int A, int B, int C, int a, int b, int c){
 68             VD ans; ans.clear();
 69             VD tmp;
 70             tmp.PB (0.0); tmp.PB (0.0); tmp.PB (0.0);
 71             double ta = A*A+B*B, tb = 2*(a*A + b*B), tc = a*a + b*b - 1;
 72             if (fabs(ta) < eps){
 73                 if (tc) return ans;
 74                 else{
 75                     if (C) return tmp;
 76                     if (!ok(0, A,B,C,a,b,c)) return ans;
 77                     ans.PB((double)a);
 78                     ans.PB((double)b);
 79                     ans.PB((double)c);
 80                     return ans;
 81                 }
 82             }
 83 
 84             double delta = tb*tb - 4 * ta * tc;
 85             if (delta < -eps) return ans;
 86             if (fabs(delta) < eps){
 87                 double t1 = -tb / 2.0 / ta;
 88                 if (!ok(t1, A,B,C,a,b,c)) return ans;
 89                 ans.PB (A*t1+a); ans.PB (B*t1+b); ans.PB (C*t1+c);
 90                 return ans;
 91             }
 92             if (delta > eps){
 93                 double t1, t[2] = {(-tb+sqrt(delta)) / 2.0/ta, (-tb-sqrt(delta)) / 2.0/ta};
 94                 int t_num = 0;
 95                 for (int i = 0; i < 2; ++ i)
 96                 if (ok(t[i],A,B,C,a,b,c)){
 97                     t1 = t[i]; ++ t_num;
 98                 }
 99                 if (t_num > 1) return tmp;
100                 else if (!t_num) return ans;
101                 ans.PB (A*t1+a); ans.PB (B*t1+b); ans.PB (C*t1+c);
102                 return ans;
103             }
104         }
105         
106 // BEGIN CUT HERE
107     public:
108     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
109     private:
110     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
111     void verify_case(int Case, const vector <double> &Expected, const vector <double> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } }
112     void test_case_0() { int Arg0 = 0; int Arg1 = 1; int Arg2 = 0; int Arg3 = -1; int Arg4 = -5; int Arg5 = -4; double Arr6[] = {}; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[0]))); verify_case(0, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
113     void test_case_1() { int Arg0 = 2; int Arg1 = 4; int Arg2 = 1; int Arg3 = -1; int Arg4 = -1; int Arg5 = 0; double Arr6[] = {0.0, 1.0, 0.5 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[0]))); verify_case(1, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
114     void test_case_2() { int Arg0 = 4; int Arg1 = 4; int Arg2 = 2; int Arg3 = 5; int Arg4 = 4; int Arg5 = 0; double Arr6[] = {0.0, 0.0, 0.0 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[0]))); verify_case(2, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
115     void test_case_3() { int Arg0 = 0; int Arg1 = 0; int Arg2 = 1; int Arg3 = 1; int Arg4 = 0; int Arg5 = 0; double Arr6[] = {0.0, 0.0, 0.0 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[0]))); verify_case(3, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
116 
117 // END CUT HERE
118 
119 };
120 
121 // BEGIN CUT HERE
122 int main()
123 {
124 //    freopen( "a.out" , "w" , stdout );    
125     ParticleCollision ___test;
126     ___test.run_test(-1);
127        return 0;
128 }
129 // END CUT HERE
View Code

 

posted @ 2013-12-17 02:04  Plumrain  阅读(196)  评论(0)  编辑  收藏  举报