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  • 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
    ",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
    ");
          return 0;
      }
    
                
    View Code
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  • 原文地址:https://www.cnblogs.com/maijing/p/4681244.html
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