NOI2013 UOJ122 向量内积

神题......

还是大神讲得比较清晰~orz

http://dffxtz.logdown.com/posts/197950-noi2013-vector-inner-product

启发题:poj3318

#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
#include<ctime>

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 int maxN=100000;
const int maxDD=100;

int N,DD,D,K;
int v[maxN+100][maxDD+10];

inline int A1(int i,int j)
  {
      if(K==3)
        {
            int t1=(j+DD-1)/DD,t2=j-(t1-1)*DD;
            return v[i][t1]*v[i][t2];
        }
      return v[i][j];
  }
inline int A2(int i,int j){return A1(j,i);}

int F[maxN+100],X[maxN+100],GX[maxN+100],FX[maxN+100],A2X[maxN+100],A1A2X[maxN+100],TX[maxN+100];
inline int work()
  {
      int i,j,k,Case;
      if(1LL*N*D<10000000LL) Case=10; else Case=2;
      while(Case--)
        {
            re(i,1,N)X[i]=rand()%2;
            
            mmst(GX,0);
            re(i,1,N)GX[1]+=X[i];
            GX[1]%=K;
            re(i,2,N)GX[i]=GX[1];
            
            re(i,1,N)FX[i]=F[i]*X[i];
            
            mmst(A2X,0);
            re(i,1,D)re(j,1,N)A2X[i]+=A2(i,j)*X[j];
            
            mmst(A1A2X,0);
            re(i,1,N)re(j,1,D)A1A2X[i]+=A1(i,j)*A2X[j];
            re(i,1,N)A1A2X[i]%=K;
            
            re(i,1,N)
              {
                  TX[i]=GX[i]-FX[i]-A1A2X[i];
                  TX[i]=(TX[i]%K+K)%K;
                  if(TX[i]!=0)
                    {
                        re(j,1,N)if(j!=i)
                          {
                              int res=0;
                              re(k,1,DD)res+=v[i][k]*v[j][k];
                              if(res%K==0){PF("%d %d\n",min(i,j),max(i,j));return 1;}
                          }
                    }
              }
        }
      return 0;
  }

int main()
  {
      freopen("meow.in","r",stdin);
      freopen("meow.out","w",stdout);
      srand(time(0));
      int i,j;
      N=gint();D=gint();K=gint();
      re(i,1,N)re(j,1,D)v[i][j]=gint()%K;
      DD=D;if(K==3)D=D*D;
      re(i,1,N)
        {
            F[i]=0;
            re(j,1,D) F[i]+=A1(i,j)*A2(j,i);
            F[i]%=K;F[i]^=1;
        }
      if(!work())PF("-1 -1\n");
      return 0;
  }

            
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posted @ 2015-07-27 21:19  maijing  阅读(236)  评论(0编辑  收藏  举报