HDU 4565So Easy! 矩阵快速幂

来源：互联网发布：查淘宝数据有哪些插件编辑：程序博客网时间：2024/05/22 08:11

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(b))^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 0x3f3f3f3fusing 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;}

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