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NOI2011 兔农

时间:2015-08-03 13:01:28      阅读:154      评论:0      收藏:0      [点我收藏+]

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http://www.lydsy.com/JudgeOnline/problem.php?id=2432

感觉是day1中最难的一题,还好出题人很良心,给了75分部分分。

还是跪拜策爷吧~Orz

http://jcvb.is-programmer.com/posts/39528.html

代码奇丑。。。。。。

技术分享
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define re(i,a,b)  for(i=a;i<=b;i++)
#define red(i,a,b) for(i=a;i>=b;i--)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=- && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==-){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-0,z=getchar());
        return (neg)?-res:res; 
    }
inline LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=- && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==-){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-0,z=getchar());
        return (neg)?-res:res; 
    }

const LL maxK=1000000;

LL N,K,P;
LL fib[6*maxK+100];
LL pos[maxK+100];
LL len[maxK+100],next[maxK+100];

inline LL gcd(LL a,LL b){return b==0 ? a : gcd(b,a%b); }
inline void extend_gcd(LL a,LL &x,LL b,LL &y)
  {
      if(b==0){x=1;y=0;return;}
      LL dx,dy;
      extend_gcd(b,dx,a%b,dy);
      x=dy;
      y=dx-a/b*dy;
  }

struct Tmatrix
  {
      int n,m;
      LL v[4][4];
      inline void clear(){n=m=0;mmst(v,0);}
      inline friend Tmatrix operator *(Tmatrix a,Tmatrix b)
        {
            int i,j,k;
            Tmatrix c;c.clear();
            c.n=a.n;c.m=b.m;
            re(i,1,c.n)re(j,1,c.m)re(k,1,a.m)c.v[i][j]=(c.v[i][j]+a.v[i][k]*b.v[k][j]%P)%P;
            return c;
        }
  };

Tmatrix A,B;

inline Tmatrix power(Tmatrix a,LL k)
  {
      int i;
      Tmatrix x,y=a;
      x.clear();x.n=x.m=a.n;re(i,1,a.n)x.v[i][i]=1;
      for(;k!=0;k>>=1){if(k&1)x=x*y;y=y*y;}
      return x;
  }

int flag[maxK+100];

int main()
  {
      /*freopen("rabbit.in","r",stdin);
      freopen("rabbit.out","w",stdout);*/
      LL i;
      N=gll();K=gll();P=gll();
      fib[1]=fib[2]=1;
      for(i=3;;i++)
        {
            fib[i]=(fib[i-1]+fib[i-2])%K;
            if(!pos[fib[i]])pos[fib[i]]=i;
            if(fib[i]==1 && fib[i-1]==1)break;
        }
      re(i,1,K-1)
        {
            LL x,y;
          if(gcd(i,K)==1)
            {
                extend_gcd(i,x,K,y);
                x=(x%K+K)%K;
                if(pos[x]==0)
                  {
                    len[i]=-1;
                                next[i]=-1;
                            }
                else
                  {
                    len[i]=pos[x];
                    next[i]=i*fib[len[i]-1]%K;
                  }
            }
          else
            len[i]=-1,next[i]=-1;
        }
      A.clear();
      A.n=A.m=3;
      A.v[1][1]=1;A.v[1][2]=1;A.v[1][3]=0;
      A.v[2][1]=1;A.v[2][2]=0;A.v[2][3]=0;
      A.v[3][1]=0;A.v[3][2]=0;A.v[3][3]=1;
      B.clear();
      B.n=B.m=3;
      B.v[1][1]=1;B.v[1][2]=0;B.v[1][3]=-1;
      B.v[2][1]=0;B.v[2][2]=1;B.v[2][3]=0;
      B.v[3][1]=0;B.v[3][2]=0;B.v[3][3]=1;
      
      int p,t;
      for(p=1;!flag[p] && next[p]!=-1;flag[p]=1,p=next[p]);
      if(next[p]!=-1)
        {
            LL lenX=0,lenY=0;Tmatrix X,Y,Z;
            X.clear();X.n=X.m=3;X.v[1][1]=X.v[2][2]=X.v[3][3]=1;
            for(t=1;t!=p;t=next[t])X=B*power(A,len[t])*X,lenX+=len[t];
            if(N>=lenX)
              {
                      Y.clear();Y.n=Y.m=3;Y.v[1][1]=Y.v[2][2]=Y.v[3][3]=1;
                      for(t=p,Y=B*power(A,len[t])*Y,lenY+=len[t],t=next[t];t!=p;t=next[t])Y=B*power(A,len[t])*Y,lenY+=len[t];
                Z=power(Y,(N-lenX)/lenY)*X;
                N=(N-lenX)%lenY;
                for(t=p;;t=next[t])
                  if(N>=len[t])
                      {
                          Z=B*power(A,len[t])*Z;
                          N-=len[t];
                      }
                    else break;
                  LL y=(Z.v[1][2]+Z.v[1][3])%P,x=(Z.v[2][2]+Z.v[2][3])%P;
                  Y.clear();
                  Y.n=Y.m=2;
                        Y.v[1][1]=1;Y.v[1][2]=1;
                        Y.v[2][1]=1;Y.v[2][2]=0;
                    Z=power(Y,N);
                    LL res=(Z.v[1][1]*y%P+Z.v[1][2]*x%P)%P;
                    cout<<(res%P+P)%P<<endl;
              }
            else
              {
                  X.clear();X.n=X.m=3;X.v[1][1]=X.v[2][2]=X.v[3][3]=1;
                  for(t=1;t!=p;t=next[t])
                    if(N>=len[t])
                      {
                          X=B*power(A,len[t])*X;
                          N-=len[t];
                      }
                    else break;
                  LL y=(X.v[1][2]+X.v[1][3])%P,x=(X.v[2][2]+X.v[2][3])%P;
                  Y.clear();
                  Y.n=Y.m=2;
                        Y.v[1][1]=1;Y.v[1][2]=1;
                        Y.v[2][1]=1;Y.v[2][2]=0;
                    Z=power(Y,N);
                    LL res=(Z.v[1][1]*y%P+Z.v[1][2]*x%P)%P;
                    cout<<(res%P+P)%P<<endl;
              }
        }
      else
        {
            LL lenX=0;Tmatrix X,Y,Z;
            X.clear();X.n=X.m=3;X.v[1][1]=X.v[2][2]=X.v[3][3]=1;
            for(t=1;t!=p;t=next[t])X=B*power(A,len[t])*X,lenX+=len[t];
            if(N>=lenX)
              {
                      LL y=(X.v[1][2]+X.v[1][3])%P,x=(X.v[2][2]+X.v[2][3])%P;
                N-=lenX;
                Y.clear();
                    Y.n=Y.m=2;
                        Y.v[1][1]=1;Y.v[1][2]=1;
                        Y.v[2][1]=1;Y.v[2][2]=0;
                    Z=power(Y,N);
                    LL res=(Z.v[1][1]*y%P+Z.v[1][2]*x%P)%P;
                    cout<<(res%P+P)%P<<endl;
                  }
                else
                  {
                      X.clear();X.n=X.m=3;X.v[1][1]=X.v[2][2]=X.v[3][3]=1;
                  for(t=1;t!=p;t=next[t])
                    if(N>=len[t])
                      {
                          X=B*power(A,len[t])*X;
                          N-=len[t];
                      }
                    else break;
                  LL y=(X.v[1][2]+X.v[1][3])%P,x=(X.v[2][2]+X.v[2][3])%P;
                  Y.n=Y.m=2;
                        Y.v[1][1]=1;Y.v[1][2]=1;
                        Y.v[2][1]=1;Y.v[2][2]=0;
                    Z=power(Y,N);
                    LL res=(Z.v[1][1]*y%P+Z.v[1][2]*x%P)%P;
                    cout<<(res%P+P)%P<<endl;
                  }
        }
      return 0;
  }
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NOI2011 兔农

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原文地址:http://www.cnblogs.com/maijing/p/4698882.html

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