BZOJ4002[JLOI2015] Meaningful string

BZOJ4002[JLOI2015] Meaningful string

Test instructions

(b+√d)/2) The integer portion of the ^n. B*b<d<10^18,n<10^18,d%4==1,b*b%4==1, modulus approximately equal to 7*10^18

Exercises

Divine questions. By some properties can draw a series: an=ban-1+ (D-b*b)/4*an-2, and this sequence of the general formula is an= ((b+√d)/2) ^n+ ((b-√d)/2) ^n, and by the topic conditions (D-B*B)/4 is a positive integer, Therefore, a can be obtained by matrix multiplication, because (b-√d)/2∈ ( -1,0], the answer is (an)-1 when and only if n is even and b*b!=d. Matrix recursion:

An-1 An-2 B 1 an An-1

*                             =

0 0 (d-b*b)/4 0 0 0

Reflection: Konjac Konjac do not know the moment multiplication does not meet the Exchange law, adjusted a long example. At the same time, because the modulus is too large, in addition to the need to use unsigned long long, multiplication is also used fast multiplication (is to use a fast power method to calculate multiplication) to prevent overflow.

Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4 #definell unsigned long long5 #defineInc (I,J,K) for (int i=j;i<=k;i++)6 #defineMoD 75284434125795769377 using namespacestd;8 9 structm{Tenll a[5][5]; OneM () {Inc (I,0,1) Inc (J,0,1) a[i][j]=0;} A }; - ll B,d,n; M St,ans; - ll Cheng (ll a,ll b) { the     if(b==0)return 0;if(b==1)returnA ll C=cheng (a,b>>1)%MoD; -     if(b&1)return((c+c)%mod+a)%mod;Else return(c+c)%MoD; - } - m Mul (M a,m b) { +M C; Inc (I,0,1) Inc (J,0,1) Inc (K,0,1) -C.a[i][j]= (C.a[i][j]+cheng (a.a[i][k],b.a[k][j]))%MoD; +     returnC; A } at m pow (M a,ll b) { -     if(b==1)returnA M C=pow (a,b>>1);if(b&1)returnMul (Mul (C,c), a);Else returnMul (c,c); - } - intMain () { -scanf"%lld%lld%lld",&b,&d,&n); -     if(n==0) {printf ("1");return 0;} inst.a[0][0]=b%mod; st.a[0][1]=2; -     if(n==1) ans=st;Else{ toans.a[0][0]=b%mod; ans.a[0][1]=1; ans.a[1][0]= (D-B*B)/4%MoD; +Ans=pow (ans,n-1); ans=Mul (St,ans); -     } the     if(b*b!=d&&! (n&1)) printf ("%lld", (ans.a[0][0]+mod-1)%mod);Elseprintf"%lld", ans.a[0][0]); *     return 0; $}

20160710

BZOJ4002[JLOI2015] Meaningful string