[数列通项 矩阵快速幂] BZOJ4002 [JLOI2015]有意义的字符串

传送门：http://blog.csdn.net/popoqqq/article/details/45148309

#include<cstdio>#include<cstdlib>#include<algorithm>#define P 7528443412579576937LLusing namespace std;typedef unsigned long long ll;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(ll &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline ll Mul(ll a,ll b){ll ret=0;for (;b;(a+=a)%=P,b>>=1)if (b&1)(ret+=a)%=P;return ret;}struct Matrix{ll a[2][2];Matrix(ll _=0,ll __=0,ll ___=0,ll ____=0) {a[0][0]=_; a[0][1]=__; a[1][0]=___; a[1][1]=____;}}A;const Matrix I(1,0,0,1);inline Matrix operator *(const Matrix &A,const Matrix &B){Matrix ret;for (int i=0;i<2;i++)for (int j=0;j<2;j++)for (int k=0;k<2;k++)(ret.a[i][j]+=Mul(A.a[i][k],B.a[k][j]))%=P;return ret;}inline Matrix Pow(Matrix a,ll b){Matrix ret=I;for (;b;a=a*a,b>>=1)if (b&1)ret=ret*a;return ret;}ll B,D,N,an;int main(){freopen("t.in","r",stdin);freopen("t.out","w",stdout);read(B); read(D); read(N);if (N==0) return printf("1\n"),0;A=Matrix(B,(((D-B*B)/4)%P+P)%P,1,0);A=Pow(A,N-1);(an=Mul(A.a[0][0],B)+Mul(A.a[0][1],2))%=P;if (B!=D*D && ~N&1) an=(an==0)?P-1:an-1;printf("%lld\n",an);return 0;}