http://www.lydsy.com/JudgeOnline/problem.php?id=3243
考虑K==2的情况
考虑把n个向量用n*d的矩阵表示，则A*AT(A的转置矩阵)中的i,j代表i和j的向量积。
考虑K==3的情况
1*1%3=1
2*2%3=1
则我们将原来矩阵中元素平方。
(A,B)^2=(a1×b1+a2×b2+a3×b3)^2
展开来就是新的向量积表示，转换之后就是像K=2一样。
详细参考:http://www.cnblogs.com/mmlz/p/4313031.html
1 #include<algorithm> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<iostream> 6 #include<time.h> 7 #define N 120005 8 #define ll long long 9 int B[120][N],A[N][120],c[N],g[N],f,T[N]; 10 int sz,s[2][N]; 11 int n,d,K; 12 int read(){ 13 int t=0,f=1;char ch=getchar(); 14 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();} 15 while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();} 16 return t*f; 17 } 18 void find(int pos){ 19 for (int i=1;i<=n;i++) 20 if (i!=pos){ 21 int res=0; 22 for (int j=1;j<=d;j++) 23 res+=A[i][j]*A[pos][j]; 24 res%=K; 25 if (res==0){ 26 printf("%d %d",std::min(i,pos),std::max(i,pos)); 27 return; 28 } 29 } 30 } 31 void mul1(){ 32 if (K==2){ 33 memset(T,0,sizeof T); 34 for (int i=1;i<=d;i++) 35 for (int j=1;j<=n;j++) 36 T[i]=(T[i]+B[i][j]*s[0][j])&1; 37 memset(s[0],0,sizeof s[0]); 38 for (int i=1;i<=n;i++) 39 for (int j=1;j<=d;j++) 40 s[0][i]=(s[0][i]+A[i][j]*T[j])&1; 41 } 42 else{ 43 memset(T,0,sizeof T); 44 for (int i=1;i<=d;i++) 45 for (int j=1;j<=d;j++) 46 for (int k=1;k<=n;k++) 47 T[i*d+j]=(T[i*d+j]+B[j][k]*B[i][k]*s[0][k])%3; 48 memset(s[0],0,sizeof s[0]); 49 for (int i=1;i<=n;i++) 50 for (int j=1;j<=d;j++) 51 for (int k=1;k<=d;k++) 52 s[0][i]=(s[0][i]+A[i][j]*A[i][k]*T[j*d+k])%3; 53 } 54 } 55 void mul2(){ 56 int sum=0; 57 for (int i=1;i<=n;i++) sum+=s[1][i]; 58 for (int i=1;i<=n;i++) 59 s[1][i]=(sum-(1-g[i])*s[1][i])%K; 60 } 61 void solve(){ 62 for (int i=1;i<=n;i++) 63 for (int j=1;j<=d;j++) 64 B[j][i]=A[i][j]; 65 for (int i=1;i<=n;i++) 66 for (int j=1;j<=d;j++){ 67 (g[i]+=A[i][j]*A[i][j])%=K; 68 if (K==3) g[i]=(g[i]*g[i])%K; 69 } 70 for (int t=1;t<=10;t++){ 71 for (int i=1;i<=n;i++) s[0][i]=rand()%2; 72 for (int i=1;i<=n;i++) s[1][i]=s[0][i]; 73 mul1();mul2(); 74 for (int i=1;i<=n;i++) 75 if (s[0][i]!=s[1][i]){ 76 find(i); 77 return; 78 } 79 } 80 } 81 int main(){ 82 srand(233); 83 n=read();d=read();K=read(); 84 for (int i=1;i<=n;i++) 85 for (int j=1;j<=d;j++) 86 (A[i][j]=read())%=K; 87 solve(); 88 return 0; 89 }