HDU 4565 So Easy! 构造矩阵（两种方法）

　A sequence S _n is defined as: 

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S _n. 
　　You, a top coder, say: So easy! 

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S _n.

Sample Input

2 3 1 20132 3 2 20132 2 1 2013

Sample Output

4144

题意：

Sn是上面那个式子的向上取整，求Sn%m

思路：

第一种方法：共轭

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>using namespace std;typedef long long LL;const int N = 2;int m;struct Matrix{    LL m[N][N];};Matrix I = {//I主对角线是1       1,0,       0,1};Matrix multi(Matrix a,Matrix b)//矩阵乘法{    Matrix c;    for(int i=0;i<N;i++)    {        for(int j=0;j<N;j++)        {            c.m[i][j] = 0;            for(int k=0;k<N;k++)                c.m[i][j] += (a.m[i][k] * b.m[k][j])%m;                c.m[i][j]%=m;        }    }    return c;}Matrix power(Matrix A,long long k)//矩阵A的k次幂（快速幂）{    Matrix ans = I,p = A;    while(k)    {        if(k&1)        {            ans = multi(ans,p);            k--;        }        k >>= 1;        p = multi(p,p);    }    return ans;}int main(){    LL a,b,n;    while(scanf("%lld%lld%lld%d",&a,&b,&n,&m)!=EOF)    {        if(n==1){printf("%lld\n",2*a%m);continue;}        if(n==0){printf("1\n");continue;}        Matrix A = {       2*a%m,(b%m-a*a%m+m)%m,       1,0};        Matrix ans = power(A,n-2);       /* for(int i=0;i<2;i++)       {           for(int j=0;j<2;j++)            cout<<ans.m[i][j]<<" ";           cout<<endl;       }*/        int c1=2*a%m;        LL c2=2*a*a%m+2*b%m;        c2%=m;       long long anss= ((ans.m[0][0]%m*c2%m)%m+ans.m[0][1]*c1%m);       anss%=m;        printf("%lld\n",anss);    }    return 0;}

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>using namespace std;typedef long long LL;const int N = 2;int m;struct Matrix{    LL m[N][N];};Matrix I = {//I主对角线是1       1,0,       0,1};Matrix multi(Matrix a,Matrix b)//矩阵乘法{    Matrix c;    for(int i=0;i<N;i++)    {        for(int j=0;j<N;j++)        {            c.m[i][j] = 0;            for(int k=0;k<N;k++)                c.m[i][j] += (a.m[i][k] * b.m[k][j])%m;                c.m[i][j]%=m;        }    }    return c;}Matrix power(Matrix A,long long k)//矩阵A的k次幂（快速幂）{    Matrix ans = I,p = A;    while(k)    {        if(k&1)        {            ans = multi(ans,p);            k--;        }        k >>= 1;        p = multi(p,p);    }    return ans;}int main(){    LL a,b,n;    while(scanf("%lld%lld%lld%d",&a,&b,&n,&m)!=EOF)    {        if(n==1){printf("%lld\n",2*a%m);continue;}        if(n==0){printf("1\n");continue;}        Matrix A = {       a%m,b%m,       1,a%m};        Matrix ans = power(A,n-1);       /* for(int i=0;i<2;i++)       {           for(int j=0;j<2;j++)            cout<<ans.m[i][j]<<" ";           cout<<endl;       }*/       long long anss= 2*((ans.m[0][0]%m*a%m)%m+ans.m[0][1]%m);        anss%=m;        printf("%lld\n",anss);    }    return 0;}