• branch=trunk inherited from [9165bd3629]
• sym-trunk inherited from [9165bd3629]

Added SRM/500/1A.cpp version [86caa30b4834e1a2]

            1  +#include <iostream>
            2  +#include <sstream>
            3  +#include <iomanip>
            4  +#include <vector>
            5  +#include <string>
            6  +#include <map>
            7  +#include <set>
            8  +#include <algorithm>
            9  +#include <numeric>
           10  +#include <iterator>
           11  +#include <functional>
           12  +#include <complex>
           13  +#include <queue>
           14  +#include <stack>
           15  +#include <cmath>
           16  +#include <cassert>
           17  +#include <cstring>
           18  +using namespace std;
           19  +typedef long long LL;
           20  +typedef complex<double> CMP;
           21  +
           22  +class MafiaGame { public:
           23  + double probabilityToLose(int N, vector <int> decisions)
           24  + {
           25  +  vector<int> vc;
           26  +  {
           27  +   map<int,int> voteCnt;
           28  +   for(int i=0; i<decisions.size(); ++i)
           29  +    voteCnt[decisions[i]] ++;
           30  +   for(map<int,int>::iterator it=voteCnt.begin(); it!=voteCnt.end(); ++it)
           31  +    vc.push_back( it->second );
           32  +  }
           33  +  const int MAX = *max_element(vc.begin(), vc.end());
           34  +  if( MAX == 1 )
           35  +   return 0.0;
           36  +
           37  +  const int GotMAX = count(vc.begin(), vc.end(), MAX);
           38  +  for(int k=GotMAX; k; k=(N-k*MAX)%k)
           39  +   if( k == 1 )
           40  +    return 1.0/GotMAX;
           41  +  return 0.0;
           42  + }
           43  +};
           44  +
           45  +// BEGIN CUT HERE
           46  +#include <ctime>
           47  +double start_time; string timer()
           48  + { ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
           49  +template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
           50  + { os << "{ ";
           51  +   for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
           52  +   os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
           53  +void verify_case(const double& Expected, const double& Received) {
           54  + bool ok = (abs(Expected - Received) < 1e-9);
           55  + if(ok) cerr << "PASSED" << timer() << endl;  else { cerr << "FAILED" << timer() << endl;
           56  + cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
           57  +#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
           58  +#define END  verify_case(_, MafiaGame().probabilityToLose(N, decisions));}
           59  +int main(){
           60  +
           61  +CASE(0)
           62  + int N = 3;
           63  + int decisions_[] = {1, 1, 1};
           64  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           65  + double _ = 1.0;
           66  +END
           67  +CASE(1)
           68  + int N = 5;
           69  + int decisions_[] = {1, 2, 3};
           70  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           71  + double _ = 0.0;
           72  +END
           73  +CASE(2)
           74  + int N = 20;
           75  + int decisions_[] = {1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 18, 19, 0};
           76  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           77  + double _ = 0.0;
           78  +END
           79  +CASE(3)
           80  + int N = 23;
           81  + int decisions_[] = {17, 10, 3, 14, 22, 5, 11, 10, 22, 3, 14, 5, 11, 17};
           82  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           83  + double _ = 0.14285714285714285;
           84  +END
           85  +/*
           86  +CASE(4)
           87  + int N = ;
           88  + int decisions_[] = ;
           89  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           90  + double _ = ;
           91  +END
           92  +CASE(5)
           93  + int N = ;
           94  + int decisions_[] = ;
           95  +   vector <int> decisions(decisions_, decisions_+sizeof(decisions_)/sizeof(*decisions_));
           96  + double _ = ;
           97  +END
           98  +*/
           99  +}
          100  +// END CUT HERE

Added SRM/500/1B.cpp version [5fcb34f3f46b9954]

            1  +#include <iostream>
            2  +#include <sstream>
            3  +#include <iomanip>
            4  +#include <vector>
            5  +#include <string>
            6  +#include <map>
            7  +#include <set>
            8  +#include <algorithm>
            9  +#include <numeric>
           10  +#include <iterator>
           11  +#include <functional>
           12  +#include <complex>
           13  +#include <queue>
           14  +#include <stack>
           15  +#include <cmath>
           16  +#include <cassert>
           17  +#include <cstring>
           18  +using namespace std;
           19  +typedef long long LL;
           20  +typedef complex<double> CMP;
           21  +
           22  +class FractalPicture { public:
           23  + double getLength(int x1, int y1, int x2, int y2)
           24  + {
           25  +  return rec(1, x1, y1, x2, y2);
           26  + }
           27  +
           28  + double rec(int i, int x1, int y1, int x2, int y2)
           29  + {
           30  +  x1 = clip<-27,+27>(x1);
           31  +  x2 = clip<-27,+27>(x2);
           32  +  y1 = clip<  0,+81>(y1);
           33  +  y2 = clip<  0,+81>(y2);
           34  +
           35  +  if( i == 501 )
           36  +   return 0;
           37  +  if( x1<=-27 && 27<=x2 && y1<=0 && 81<=y2 )
           38  +   return 81 + (500-i)*54;
           39  +  if( x2<=-27 || 27<=x1 || y2<=0 || 81<=y1 )
           40  +   return 0;
           41  +  double L0 = (x1<=0 && 0<=x2 ? clip<0,54>(y2)-clip<0,54>(y1) : 0);
           42  +  x1=x1*3, x2=x2*3, y1=(y1-54)*3, y2=(y2-54)*3;
           43  +  double L1 = rec(i+1, x1, y1, x2, y2);
           44  +  double L2 = rec(i+1, y1, x1, y2, x2);
           45  +  double L3 = rec(i+1, y1, -x2, y2, -x1);
           46  +  return L0 + (L1+L2+L3)/3;
           47  + }
           48  +
           49  + template<int m, int M>
           50  + int clip(int v) { return max(m, min(v, M)); }
           51  +};
           52  +
           53  +// BEGIN CUT HERE
           54  +#include <ctime>
           55  +double start_time; string timer()
           56  + { ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
           57  +template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
           58  + { os << "{ ";
           59  +   for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
           60  +   os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
           61  +void verify_case(const double& Expected, const double& Received) {
           62  + bool ok = (abs(Expected - Received) < 1e-9);
           63  + if(ok) cerr << "PASSED" << timer() << endl;  else { cerr << "FAILED" << timer() << endl;
           64  + cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
           65  +#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
           66  +#define END  verify_case(_, FractalPicture().getLength(x1, y1, x2, y2));}
           67  +int main(){
           68  +
           69  +CASE(0)
           70  + int x1 = -1;
           71  + int y1 = 0;
           72  + int x2 = 1;
           73  + int y2 = 53;
           74  + double _ = 53.0;
           75  +END
           76  +CASE(1)
           77  + int x1 = 1;
           78  + int y1 = 1;
           79  + int x2 = 10;
           80  + int y2 = 10;
           81  + double _ = 0.0;
           82  +END
           83  +CASE(2)
           84  + int x1 = -10;
           85  + int y1 = 54;
           86  + int x2 = 10;
           87  + int y2 = 55;
           88  + double _ = 21.0;
           89  +END
           90  +CASE(3)
           91  + int x1 = 14;
           92  + int y1 = 45;
           93  + int x2 = 22;
           94  + int y2 = 54;
           95  + double _ = 2999.0;
           96  +END
           97  +CASE(4)
           98  + int x1 = -28;
           99  + int y1 = 44;
          100  + int x2 = -16;
          101  + int y2 = 62;
          102  + double _ = 6992.0;
          103  +END
          104  +CASE(5)
          105  + int x1 = 0;
          106  + int y1 = 80;
          107  + int x2 = 1;
          108  + int y2 = 81;
          109  + double _ = 332;
          110  +END
          111  +}
          112  +// END CUT HERE

Added SRM/500/1C.cpp version [45a3476fb0482db3]

            1  +#include <iostream>
            2  +#include <sstream>
            3  +#include <iomanip>
            4  +#include <vector>
            5  +#include <string>
            6  +#include <map>
            7  +#include <set>
            8  +#include <algorithm>
            9  +#include <numeric>
           10  +#include <iterator>
           11  +#include <functional>
           12  +#include <complex>
           13  +#include <queue>
           14  +#include <stack>
           15  +#include <cmath>
           16  +#include <cassert>
           17  +#include <cstring>
           18  +using namespace std;
           19  +typedef long long LL;
           20  +typedef complex<double> CMP;
           21  +
           22  +static const int MODVAL = 500500573;
           23  +struct mint
           24  +{
           25  + int val;
           26  + mint():val(0){}
           27  + mint(int x):val(x%MODVAL) {}
           28  + mint(LL  x):val(x%MODVAL) {}
           29  +};
           30  +mint operator+(mint x, mint y) { return x.val+y.val; }
           31  +mint operator-(mint x, mint y) { return x.val-y.val+MODVAL; }
           32  +mint operator*(mint x, mint y) { return LL(x.val)*y.val; }
           33  +mint POW(mint x, int e) {
           34  + mint v = 1;
           35  + for(;e;x=x*x,e>>=1)
           36  +  if(e&1)
           37  +   v=v*x;
           38  + return v;
           39  +}
           40  +mint operator/(mint x, mint y) { return x * POW(y, MODVAL-2); }
           41  +
           42  +class ProductAndSum { public:
           43  + int getSum(int p2, int p3, int p5, int p7, int S)
           44  + {
           45  +  // Precomputing n!, 1/n!, and 111...111
           46  +  vector<mint> FACT(2501), FACT_INV(2501), ONES(2501);
           47  +  FACT[0] = FACT_INV[0] = 1;
           48  +  ONES[0] = 0;
           49  +  for(int n=1; n<FACT.size(); ++n)
           50  +  {
           51  +   FACT[n] = FACT[n-1] * n;
           52  +   FACT_INV[n] = 1 / FACT[n];
           53  +   ONES[n] = ONES[n-1]*10 + 1;
           54  +  }
           55  +
           56  +  mint answer = 0;
           57  +
           58  +  // Exhaustive search over the number of 4,6,8,9s : at most 33*50*50*100 < 8M iterations
           59  +  for(int v8=0; v8*3<=p2; ++v8)
           60  +  for(int v4=0; v4*2+v8*3<=p2; ++v4)
           61  +  for(int v9=0; v9*2<=p3; ++v9)
           62  +  for(int v6=0; v6+v4*2+v8*3<=p2 && v6+v9*2<=p3; ++v6)
           63  +  {
           64  +   // then, the number of 1,2,3s are computable
           65  +   int v[] = {-1, -1, p2-v8*3-v4*2-v6, p3-v9*2-v6, v4, p5, v6, p7, v8, v9};
           66  +   {
           67  +    int v1 = S;
           68  +    for(int i=2; i<=9; ++i)
           69  +     v1 -= i * v[i];
           70  +    if( v1 < 0 )
           71  +     continue;
           72  +    v[1] = v1;
           73  +   }
           74  +   int N = accumulate(v+1, v+10, 0);
           75  +
           76  +   // Now, let's compute the sum of all integers, who have N digits and #d = v[d], in constant time!
           77  +
           78  +   // It can be calculated as follows:
           79  +   //   Q: how many of the integers have, say "1" in the last digit?
           80  +   //   A: it is, of course, C(N-1, v[1]-1, v[2], ..., v[9])
           81  +   //      where C(k,a,b,..) is the # of ways of spliting k elements into groups of size a, b, ...
           82  +   // So, the sum of the least digits of the integers are
           83  +   //   X = \Sigma_{d=1..9} (d * C(N-1, v[1], v[2], ..., v[d]-1, ..., v[9]))
           84  +   // By the same argument, the sum of the second-last digits (10-no-kurai in japanese) is
           85  +   //   10 * X
           86  +   // For hundreds, the sum is 100*X, and so on. Eventualy, the sum of whole integers is
           87  +   //   t = (1+10+...+10^N) * X
           88  +   //     = 111...111 * \Sigma_{d=1..9} (d * C(N-1, v[1], v[2], ..., v[d]-1, ..., v[9]))
           89  +   // Let us simplify the formula by using C(k,a,b,...) = k! / a!b!...
           90  +   //   t = 111...111 * \Sigma_{d=1..9} (d * (N-1)! / v[1]! / ... / v[9]! * v[d])
           91  +   //     = 111...111 * (N-1)! / v[1]! / ... / v[9]! * \Sigma_{d=1..9} (d*v[d])
           92  +   //     = 111...111 * (N-1)! / v[1]! / ... / v[9]! * S
           93  +   // That's it!
           94  +
           95  +   mint t = ONES[N] * FACT[N-1] * S;
           96  +   for(int i=1; i<=9; ++i)
           97  +    t = t * FACT_INV[v[i]];
           98  +   answer = answer + t;
           99  +  }
          100  +  return answer.val;
          101  + }
          102  +};
          103  +
          104  +// BEGIN CUT HERE
          105  +#include <ctime>
          106  +double start_time; string timer()
          107  + { ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
          108  +template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
          109  + { os << "{ ";
          110  +   for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
          111  +   os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
          112  +void verify_case(const int& Expected, const int& Received) {
          113  + bool ok = (Expected == Received);
          114  + if(ok) cerr << "PASSED" << timer() << endl;  else { cerr << "FAILED" << timer() << endl;
          115  + cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
          116  +#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
          117  +#define END  verify_case(_, ProductAndSum().getSum(p2, p3, p5, p7, S));}
          118  +int main(){
          119  +
          120  +CASE(0)
          121  + int p2 = 2;
          122  + int p3 = 0;
          123  + int p5 = 0;
          124  + int p7 = 0;
          125  + int S = 4;
          126  + int _ = 26;
          127  +END
          128  +CASE(1)
          129  + int p2 = 0;
          130  + int p3 = 0;
          131  + int p5 = 0;
          132  + int p7 = 0;
          133  + int S = 10;
          134  + int _ = 110109965;
          135  +END
          136  +CASE(2)
          137  + int p2 = 2;
          138  + int p3 = 0;
          139  + int p5 = 0;
          140  + int p7 = 0;
          141  + int S = 5;
          142  + int _ = 610;
          143  +END
          144  +CASE(3)
          145  + int p2 = 1;
          146  + int p3 = 1;
          147  + int p5 = 1;
          148  + int p7 = 1;
          149  + int S = 10;
          150  + int _ = 0;
          151  +END
          152  +CASE(4)
          153  + int p2 = 5;
          154  + int p3 = 5;
          155  + int p5 = 5;
          156  + int p7 = 5;
          157  + int S = 100;
          158  + int _ = 61610122;
          159  +END
          160  +CASE(5)
          161  + int p2 = 100;
          162  + int p3 = 100;
          163  + int p5 = 100;
          164  + int p7 = 100;
          165  + int S = 2500;
          166  + int _ = -1;
          167  +END
          168  +/*
          169  +CASE(6)
          170  + int p2 = ;
          171  + int p3 = ;
          172  + int p5 = ;
          173  + int p7 = ;
          174  + int S = ;
          175  + int _ = ;
          176  +END
          177  +*/
          178  +}
          179  +// END CUT HERE