hdu 4565 So Easy!

原题目
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.

解释：给出一个公式：
要求小数点后一但有数往上进位（只保留整数位上的数）；求出Sn

题解：简单的矩阵快速幂，可以直接套模板过；
本题主要事考察能否在短时间内观察出题目所给的公式的规律：通过找共轭来规避小数点后的计算；
建立矩阵 e ：{2*a | -(a^2-b)}
：{1 | 0 }
矩阵 c {2*a}
{1}
本题要求的便是e^(n-1)*矩阵 c的结果；
给出代码

#include <stdio.h>#include <math.h>#include <stdlib.h>#include <time.h>#include <string.h>#define maxn 100007using namespace std;typedef unsigned long long ull;typedef long long ll;ll a,b,n,m;struct mat{    ll val[5][5];};mat matmult(mat a,mat b){    mat ms;    memset(ms.val,0,sizeof(ms.val));    for(int i=1;i<=2;i++)    {        for(int j=1;j<=2;j++)        {            for(int k=1;k<=2;k++)            {                ms.val[i][j]+=a.val[i][k]*b.val[k][j];                ms.val[i][j]%=m;            }        }    }    return ms;}mat matpow(mat a,ll k){    mat ans;    memset(ans.val,0,sizeof(ans.val));    for(int i=1;i<=2;i++)        ans.val[i][i]=1;    while(k)    {        if(k&1)            ans=matmult(a,ans);        a=matmult(a,a);        k>>=1;    }    return ans;}int main(){    while(scanf("%lld%lld%lld%lld",&a,&b,&n,&m)==4)    {        mat c;        mat e;        e.val[1][1]=2*a%m;        e.val[1][2]=-(a*a-b)%m;        e.val[2][1]=1;        e.val[2][2]=0;        if(n==1)        {            printf("%lld\n",2*a%m);            continue;        }        e=matpow(e,n-1);        printf("%lld\n",((e.val[1][1]*2*a+e.val[1][2]*2)%m+m)%m);    }    return 0;}