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=1010;
int n,m,k;
struct Matrix{double a[Maxn];}Ori,Bas,Ans;
inline Matrix operator * (Matrix A,Matrix B)
{
    Matrix C; memset(C.a,0,sizeof(C.a));
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) C.a[i]+=A.a[j]*B.a[(i-j+n+1)%n==0?n:(i-j+n+1)%n];
    return C;
}
inline Matrix Pow(Matrix Base,int Exp)
{
    Matrix Ret; memset(Ret.a,0,sizeof(Ret.a)); Ret.a[1]=1;
    while (Exp)
    {
        if (Exp&1) Ret=Ret*Base;
        Base=Base*Base; Exp>>=1;
    }
    return Ret;
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for (int i=1;i<=n;i++) scanf("%lf",&Ori.a[i]);
    Bas.a[1]=(double)(m-1)/(double)m;
    Bas.a[2]=1/(double)m;
    Bas=Pow(Bas,k);
    Ans=Bas*Ori;
    for (int i=1;i<=n;i++) printf("%.3lf
",Ans.a[i]);
    return 0;
}

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