5451 Best Solver 构造共轭复根求递推矩阵&广义斐波那契循环节降幂

题意：有式子$y=(5+2\sqrt6)^{1+2^x}$.
给定 $x$和$M$ ($0 \leq x < 2^{32}$) 求$y(x)$的整数部分$mod$ $M$的值是多少（向下取整）

推矩阵资料：http://blog.csdn.net/crazy______/article/details/9021169
广义斐波那契循环节：http://blog.csdn.net/acdreamers/article/details/25616461
构造类斐波那契数列矩阵:http://blog.csdn.net/acdreamers/article/details/8994222

。。真是。。神。。
可以根据式子推出矩阵 f(n+1)=10*f(n)-f(n-1)
注意f(1)=10 f(2)=2

//author: CHC
//First Edit Time:    2015-09-20 20:31
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#include <limits>
using namespace std;
typedef long long LL;
const int MAXN=1e+4;
const int MAXM=1e+5;
const int INF = numeric_limits<int>::max();
const LL LL_INF= numeric_limits<LL>::max();
const int N = 2;
void matric_mul(LL A[][N],LL B[][N],LL C[][N],LL mod){
    LL tmp[N][N];
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            tmp[i][j]=0;
            for(int k=0;k<N;k++){
                tmp[i][j]=((tmp[i][j]+A[i][k]*B[k][j])%mod+mod)%mod;
            }
        }
    }
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            C[i][j]=tmp[i][j];
        }
    }
}
LL q_pow(LL n,LL mod){
    LL tmp[][N]={{0,-1},{1,10}};
    LL ans[N][N];
    memset(ans,0,sizeof(ans));
    for(int i=0;i<N;i++)ans[i][i]=1;
    while(n){
        if(n&1)matric_mul(ans,tmp,ans,mod);
        n>>=1;
        matric_mul(tmp,tmp,tmp,mod);
    }
    //printf("%I64d %I64d\n%I64d %I64d\n",ans[0][0],ans[0][1],ans[1][0],ans[1][1]);
    //puts("");
    //printf("%I64d %I64d\n",ans[1][1],ans[0][1]);
    //return ans[1][1]%mod;
    return ((((ans[1][1]*10)%mod+mod)%mod+((ans[0][1]*2)%mod+mod)%mod)%mod+mod)%mod;
}
LL q_pow1(LL base,LL n,LL mod){
    LL ans=1;
    while(n){
        if(n&1)ans=ans*base%mod;
        n>>=1;
        base=base*base%mod;
    }
    return ans;
}
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--){
        LL x;
        int mod;
        scanf("%I64d%d",&x,&mod);
        LL p=(LL)(mod+1)*(mod-1);
        LL r=q_pow1(2,x,p);
        //printf("r:%I64d p:%I64d\n",r,p);
        LL res=q_pow(r,mod);
        //printf("%I64d\n",res);
        printf("Case #%d: ",++cas);
        if(!res){
            printf("%d\n",mod-1);
        }
        else {
            printf("%I64d\n",res-1);
        }
        //printf("%I64d\n",q_pow(x,mod));
        //LL r=q_pow1(2LL,(LL)x,(LL)(mod-1)*(mod+1));
        //printf("%I64d\n",q_pow(r,(LL)mod));
    }
    return 0;
}