数学（矩阵乘法）：HDU 4565 So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3804 Accepted Submission(s): 1251


[align=left]Problem Description[/align]
　　A sequence Sn 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 Sn.
　　You, a top coder, say: So easy!



[align=left]Input[/align]
　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the end of file.

[align=left]Output[/align]
　　For each the case, output an integer Sn.

[align=left]Sample Input[/align]

2 3 1 2013
2 3 2 2013
2 2 1 2013

[align=left]Sample Output[/align]

4
14
4
　　这道题还是有点门路的，需要一些巧法。
　　由于(a-1)2< b < a2，可以得出a-1<sqrt(b)<a,0<a-sqrt(b)<1。
　　这时构建E(n)=(a+sqrt(b))^n+(a-sqrt(b))^n,发现E(n)是整数，而且(a-sqrt(b))^n小于1，那么(a+sqrt(b))^n向上取整就是E(n)。
　　通过推导可以得出E(0)=2,E(1)=2*a,E(n)=2*a*E(n-1)-(a*a-b)*E(n-2),用矩阵乘法快速求出即可。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
const int maxn=210;
int a,b,n,m;
struct Array{
int a[maxn],L;
void Init(int x){L=x;memset(a,0,sizeof(a));}
int *operator[](int x){return &a[(x-1)*L];}
};
struct Matrix{
int R,C;
Array mat;
void Init(int r,int c){mat.Init(c);R=r;C=c;}
int *operator[](int x){return mat[x];}
friend Matrix operator*(Matrix a,Matrix b){
Matrix c;c.Init(a.R,b.C);
for(int i=1;i<=a.R;i++)
for(int j=1;j<=b.C;j++)
for(int k=1;k<=a.C;k++)
(c[i][j]+=1ll*a[i][k]*b[k][j]%m)%=m;
return c;
}
friend Matrix operator^(Matrix a,int k){
Matrix c;c.Init(a.R,a.C);
for(int i=1;i<=a.R;i++)c[i][i]=1;
while(k){if(k&1)c=c*a;k>>=1;a=a*a;}
return c;
}
}A,B;

int main(){
while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF){
A.Init(2,2);
A[1][1]=2*a%m;A[1][2]=(b-1ll*a*a%m+m)%m;
A[2][1]=1;A[2][2]=0;

B.Init(2,1);
B[1][1]=2*a%m;
B[2][1]=2;
A=A^(n-1);B=A*B;
printf("%d\n",B[1][1]);
}
return 0;
}