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

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "FIELDDiagrams.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int maxint = ;

 int64 d[][];

 class FIELDDiagrams
 {
     public:
         long long countDiagrams(int f){
             ++ f;
             for (int i = ; i < f; ++ i)
                 d[][i] = ;
             for (int i = ; i < f; ++ i){
                 for (int j = ; j <= f-i; ++ j){
                     d[i][j] = ;
                     for (int k = j; k <= f-i+; ++ k)
                         d[i][j] += d[i-][k];
                 }
             }
             return d[f-][] + d[f-][] - ;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); }
     private:
     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(); }
     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; } }
     void test_case_0() { int Arg0 = ; long long Arg1 = 4LL; verify_case(, Arg1, countDiagrams(Arg0)); }
     void test_case_1() { int Arg0 = ; long long Arg1 = 13LL; verify_case(, Arg1, countDiagrams(Arg0)); }
     void test_case_2() { int Arg0 = ; long long Arg1 = 131LL; verify_case(, Arg1, countDiagrams(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     FIELDDiagrams ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

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

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "ParticleCollision.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int maxint = ;

 bool ok(double t, int A, int B, int C, int a, int b, int c)
 {
     double x2 = A*t + a, y2 = B*t + b, z = C*t + c;
     double x1 = cos(z*atan(1.0)*), y1 = sin(z*atan(1.0)*);
     if (fabs(x1-x2) < eps && fabs(y1-y2) < eps) return ;
     return ;
 }

 class ParticleCollision
 {
     public:
         vector <double> collision(int A, int B, int C, int a, int b, int c){
             VD ans; ans.clear();
             VD tmp;
             tmp.PB (0.0); tmp.PB (0.0); tmp.PB (0.0);
             double ta = A*A+B*B, tb = *(a*A + b*B), tc = a*a + b*b - ;
             if (fabs(ta) < eps){
                 if (tc) return ans;
                 else{
                     if (C) return tmp;
                     if (!ok(, A,B,C,a,b,c)) return ans;
                     ans.PB((double)a);
                     ans.PB((double)b);
                     ans.PB((double)c);
                     return ans;
                 }
             }

             double delta = tb*tb -  * ta * tc;
             if (delta < -eps) return ans;
             if (fabs(delta) < eps){
                 double t1 = -tb / 2.0 / ta;
                 if (!ok(t1, A,B,C,a,b,c)) return ans;
                 ans.PB (A*t1+a); ans.PB (B*t1+b); ans.PB (C*t1+c);
                 return ans;
             }
             if (delta > eps){
                 double t1, t[] = {(-tb+sqrt(delta)) / 2.0/ta, (-tb-sqrt(delta)) / 2.0/ta};
                 int t_num = ;
                 for (int i = ; i < ; ++ i)
                 if (ok(t[i],A,B,C,a,b,c)){
                     t1 = t[i]; ++ t_num;
                 }
                 if (t_num > ) return tmp;
                 else if (!t_num) return ans;
                 ans.PB (A*t1+a); ans.PB (B*t1+b); ans.PB (C*t1+c);
                 return ans;
             }
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     private:
     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(); }
     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; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = -; int Arg4 = -; int Arg5 = -; double Arr6[] = {}; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[]))); verify_case(, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = -; int Arg4 = -; int Arg5 = ; double Arr6[] = {0.0, 1.0, 0.5 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[]))); verify_case(, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; double Arr6[] = {0.0, 0.0, 0.0 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[]))); verify_case(, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; double Arr6[] = {0.0, 0.0, 0.0 }; vector <double> Arg6(Arr6, Arr6 + (sizeof(Arr6) / sizeof(Arr6[]))); verify_case(, Arg6, collision(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     ParticleCollision ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE