HDU 4565 So Easy! 矩阵快速幂

设(a+sqrt(b))^n为(Xn + Yn*sqrt(b))，那么显然有(a+sqrt(b))^(n+1) 为 (a*Xn + b*Yn + (aYn+Xn)*sqrt(b))。

那么显然有(a+sqrt(b))的Xn，Yn可以表示为 :

然后又会发现，(a-sqrt(6))^n可以表示为：

那么会发现(a+sqrt(b))^n = (a+sqrt(b))^n + (a-sqrt(b))^n - (a-sqrt(b))^n = Xn+Yn*sqrt(b) +Xn-Yn*sqrt(b) -(a-sqrt(b))^n = 2*Xn - (a-sqrt(b))^n。

又由题意得a-sqrt(b)∈(0,1)，切最终答案向上取整，所以可得最终答案为2×Xn。

去年长沙网赛的题，这种题都不能1Y还怎么玩？

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000")
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f

using namespace std;

const int MAXN = 61;

struct MAT
{
    int row,col;
    int mat[MAXN][MAXN];

    void Init(int R,int C,int val)
    {
        row = R,col = C;

        for(int i = 1;i <= row; ++i)
            for(int j = 1;j <= col; ++j)
                    mat[i][j] = (i == j ? val : 0);
    }

    MAT Multi(MAT c,int MOD)
    {
        MAT tmp;
        tmp.Init(this->row,c.col,0);

        int i,j,k;
        for(k = 1;k <= this->col; ++k)
            for(i = 1;i <= tmp.row; ++i)
                for(j = 1;j <= tmp.col; ++j)
                    (tmp.mat[i][j] += (this->mat[i][k]*c.mat[k][j])%MOD) %= MOD;
        return tmp;
    }

    MAT Quick(int n,int MOD)
    {
        MAT res,tmp = *this;

        res.Init(row,col,1);

        while(n)
        {
            if(n&1)
                res = res.Multi(tmp,MOD);
            tmp = tmp.Multi(tmp,MOD);
            n >>= 1;
        }

        return res;
    }

    void Output()
    {
        cout<<"         ****************        "<<endl;
        int i,j;
        for(i = 1;i <= row; ++i)
        {
            for(j = 1;j <= col; ++j)
                    printf("%3d ",mat[i][j]);
            puts("");
        }
        cout<<"         &&&&&&&&&&&&&       "<<endl;
    }

};

int main()
{
    int a,b,n,m;

    MAT A,B;

    freopen("data1.in","r",stdin);

    while(scanf("%d %d %d %d",&a,&b,&n,&m) != EOF)
    {
        a %= m,b %= m;
        A.Init(2,1,0);
        B.Init(2,2,0);

        B.mat[1][1] = a;
        B.mat[1][2] = b;

        B.mat[2][1] = 1;
        B.mat[2][2] = a;

        A.mat[1][1] = a;
        A.mat[2][1] = 1;
        B = B.Quick(n-1,m);
        B = B.Multi(A,m);

        printf("%d\n",(2*B.mat[1][1])%m);
    }
    return 0;
}

HDU 4565 So Easy! 矩阵快速幂

原文：http://blog.csdn.net/zmx354/article/details/44056817