BZOJ 4204 && BZOJ 2510 循环矩阵

n^3logn非常显然。所以要用一种因为这个矩阵是一个循环矩阵，所以只要知道第一行就可以知道所有行了。 C[i][j]=C[i-1][j-1];

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int Maxn=;
 int n,m,k;
 struct Matrix{double a[Maxn];}Ori,Bas,Ans;
 inline Matrix operator * (Matrix A,Matrix B)
 {
     Matrix C; memset(C.a,,sizeof(C.a));
     for (int i=;i<=n;i++)
         for (int j=;j<=n;j++) C.a[i]+=A.a[j]*B.a[(i-j+n+)%n==?n:(i-j+n+)%n];
     return C;
 }
 inline Matrix Pow(Matrix Base,int Exp)
 {
     Matrix Ret; memset(Ret.a,,sizeof(Ret.a)); Ret.a[]=;
     while (Exp)
     {
         if (Exp&) Ret=Ret*Base;
         Base=Base*Base; Exp>>=;
     }
     return Ret;
 }
 int main()
 {
     scanf("%d%d%d",&n,&m,&k);
     for (int i=;i<=n;i++) scanf("%lf",&Ori.a[i]);
     Bas.a[]=(double)(m-)/(double)m;
     Bas.a[]=/(double)m;
     Bas=Pow(Bas,k);
     Ans=Bas*Ori;
     for (int i=;i<=n;i++) printf("%.3lf\n",Ans.a[i]);
     return ;
 }
 

C++

C[1][i]=A[1][k]*B[k][i]; 就可以转换为 C.a[1][i]=A[1][k]*B[1][i-k+1];

类似题目:POJ 3150 HDU 2276 HDU 4576