hdu5451Best Solver=矩阵快速幂+广义斐波拉契

Problem Description
The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart.

It is known that $y=(5+2\sqrt6)^{1+2^x}.$
For a given integer x (0≤x<2^32) and a given prime number M (M≤46337), print [y]%M. ([y] means the integer part of y)

Input
An integer T (1< T≤1000), indicating there are T test cases.
Following are T lines, each containing two integers x and M, as introduced above.

Output
The output contains exactly T lines.
Each line contains an integer representing [y]%M.

Sample Input

7
0 46337
1 46337
3 46337
1 46337
21 46337
321 46337
4321 46337

Sample Output

Case #1: 97
Case #2: 969
Case #3: 16537
Case #4: 969
Case #5: 40453
Case #6: 10211
Case #7: 17947

Source
2015 ACM/ICPC Asia Regional Shenyang Online

参考：http://blog.youkuaiyun.com/xtulollipop/article/details/52382791
同样的做法：然后就可以去找矩阵的循环节，可以暴力扫，也可以用结论：
http://blog.youkuaiyun.com/xtulollipop/article/details/52373948
然后就简单了。。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define pi acos(-1.0)
#define EPS 1e-6    //log(x)
#define e exp(1.0); //2.718281828
//#define mod 1000000007
#define INF 0x7fffffff
#define inf 0x3f3f3f3f
#pragma comment(linker,"/STACK:102400000,102400000")
typedef long long LL;

#define debug(x) cout<<x<<endl;
#define debug2(x) cout<<x<<" ";

//#define MOD 10000007
LL MOD;
struct Mat{
    int n,m;
    LL mat[9][9];
};
Mat operator *(Mat a,Mat b){
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    c.n = a.n,c.m = b.m;

    for(int i=1;i<=a.n;i++){
        for(int j=1;j<=b.m;j++){
            for(int k=1;k<=a.m;k++){
                c.mat[i][j] += (a.mat[i][k]*b.mat[k][j])%MOD;
                c.mat[i][j] %= MOD;
            }
        }
    }
    return c;
}
Mat operator +(Mat a,Mat b){
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    c.n = a.n,c.m = a.m;

    for(int i=1;i<=a.n;i++){
        for(int j=1;j<=a.m;j++){
            c.mat[i][j] = (a.mat[i][j]+b.mat[i][j])%MOD;
        }
    }
    return c;
}
Mat operator ^(Mat a,LL k){
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    c.n = a.n,c.m = a.n;
    for(int i=1;i<=a.n;i++)c.mat[i][i] = 1;

    while(k){
        if(k&1){
            c = c*a;
        }
        a = a*a;
        k>>=1;
    }
    return c;
}
void out(Mat a){
    for(int i=1;i<=a.n;i++){
        for(int j=1;j<=a.m;j++){
            printf(j==a.m? "%I64d\n":"%I64d ",a.mat[i][j]);
        }
    }
}
LL quickPow(LL x, LL n, LL mm)
{
    LL a = 1;
    while (n)
    {
        a *= n&1 ? x : 1;
        a %= mm;
        n >>= 1 ;
        x *= x;
        x %= mm;
    }
    return a;
}
int main()
{
    int T_T;
    scanf("%d",&T_T);
    LL x,m;
    int cas=0;
    while(T_T--){
        scanf("%I64d %I64d",&x,&m);
        printf("Case #%d: ",++cas);
        MOD=m;
        LL tempmod=m*m-1;
        LL n=quickPow(2,x,tempmod)+1;
        if(n==0){
            printf("1\n");
            continue;
        }
        else if(n==1){
            printf("9\n");
            continue;
        }
        Mat pp;
        pp.n=pp.m=2;
        pp.mat[1][1]=5%MOD;
        pp.mat[1][2]=12%MOD;
        pp.mat[2][1]=2;
        pp.mat[2][2]=5%MOD;

        Mat A0;
        A0.n=2,A0.m=1;
        A0.mat[1][1]=5%MOD;
        A0.mat[2][1]=2;

        Mat ans=pp^(n-1);
        ans=ans*A0;

        printf("%I64d\n",(2*ans.mat[1][1]-1+MOD)%MOD);
    }
    return 0;
}


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                   `=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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