HDU 5451 Best Solver

没有意识到循环节最大是M^2，用域Z_M下二阶可逆矩阵群的最大循环节(M^2-1)*(M^2-M)来做，考虑大数乘法，矩阵乘法，做到吐血。将代码贴在下方留作纪念。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define ms(arr,a) memset(arr,a,sizeof arr)
long long P;
long long M;
long long x;
long long modM(long long x)
{
    if(x<0)return x%M+M;
    else return x%M;
}
struct matrix
{
    long long a,b,c,d;
    matrix(){}
    matrix(long long m,long long n,long long p,long long q):a(m),b(n),c(p),d(q){}
    matrix(long long x):a(x),b(0),c(0),d(x){}
};
matrix solve(matrix A,matrix B)
{
    matrix ret;
    ret.a=modM(A.a*B.a+A.b*B.c);
    ret.b=modM(A.a*B.b+A.b*B.d);
    ret.c=modM(A.c*B.a+A.d*B.c);
    ret.d=modM(A.c*B.b+A.d*B.d);
    return ret;
}
long long solve(long long x,long long y)
{
    long long ret=0;
    while(x)
    {
        if(x&1)ret=(ret+y)%P;
        y=(2*y)%P;
        x>>=1;
    }
    return ret;
}
template<typename T>
T quick_pow(T a,long long n)
{
    T ret(1);
    while(n)
    {
        if(n&1)ret=solve(ret,a);
        a=solve(a,a);
        n>>=1;
    }
    return ret;
}
int main()
{
    int T;scanf("%d",&T);
    rep(Case,1,T)
    {
        scanf("%lld%lld",&x,&M);
        P=(M*M-1)*(M*M-M);
        long long N=(quick_pow(2LL,x)+1)%P;
        //printf("%lld
",N);
        matrix tmp=(matrix){10,-1,1,0};
        tmp=quick_pow(tmp,N);
        printf("Case #%d: %lld
",Case,modM(tmp.c*10+tmp.d*2-1));
    }
}