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<)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(),c(),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=;
     while(x)
     {
         if(x&)ret=(ret+y)%P;
         y=(*y)%P;
         x>>=;
     }
     return ret;
 }
 template<typename T>
 T quick_pow(T a,long long n)
 {
     T ret();
     while(n)
     {
         if(n&)ret=solve(ret,a);
         a=solve(a,a);
         n>>=;
     }
     return ret;
 }
 int main()
 {
     int T;scanf("%d",&T);
     rep(Case,,T)
     {
         scanf("%lld%lld",&x,&M);
         P=(M*M-)*(M*M-M);
         long long N=(quick_pow(2LL,x)+)%P;
         //printf("%lld\n",N);
         matrix tmp=(matrix){,-,,};
         tmp=quick_pow(tmp,N);
         printf("Case #%d: %lld\n",Case,modM(tmp.c*+tmp.d*-));
     }
 }