标签：$$ SM space 黄金分割 void struct 素数 signed pac
1 inline int Fib(int n) 2 { 3 if (n == 1 || n == 2) 4 { 5 return 1; 6 } 7 return Fib(n - 1) + Fib(n - 2); 8 }
1 inline int Fib(int n) 2 { 3 F[1] = F[2] = 1; 4 for(int i = 3 ; i <= n ; ++ i) 5 F[i] = F[i - 1] + F[i - 2]; 6 return F[n]; 7 }
复杂度分析 :O(n)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define inc(i) (++ i) 5 #define int long long 6 using namespace std; 7 const int Mod = 1000000000 + 7; 8 struct JX 9 { 10 int a[3][3]; 11 friend JX operator * (JX a , JX b) 12 { 13 JX c; 14 memset(c.a , 0 , sizeof(c.a)); 15 for(int i = 1 ; i <= 2 ; inc(i)) 16 for(int j = 1 ; j <= 2 ; inc(j)) 17 for(int k = 1 ; k <= 2 ; inc(k)) 18 c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j]) % Mod; 19 return c; 20 } 21 }Ans , A; 22 int n; 23 inline void KSM() 24 { 25 int p = n - 2; 26 while(p) 27 { 28 if(p & 1) Ans = Ans * A; 29 p >>= 1 , A = A * A; 30 } 31 } 32 signed main() 33 { 34 scanf("%ld" , &n); 35 if(n <= 2) putchar(49); 36 else 37 { 38 Ans.a[1][1] = A.a[1][1] = 1;//直接跳到n = 2时的矩阵 39 Ans.a[1][2] = A.a[1][2] = 1; 40 Ans.a[2][1] = A.a[2][1] = 1; 41 KSM(); 42 printf("%lld" , Ans.a[1][1]); 43 } 44 return 0; 45 }
1#include<iostream>2#include<cstdio>3#include<cctype>4#include<cstring>5#include<cmath>6#include<algorithm>7#defineintlonglong8#defineinc(i)(++i)9#definedec(i)(--i)10usingnamespacestd;11constintN=100000+7;12intn,Prime[N+7],tot,Pri[N],Cnt[N],cnt,TOT;13intFactor[N],P;14boolNo_Prime[N+7];15structJX16{17inta[4][4];18friendJXoperator*(JXa,JXb)19{20JXc;21memset(c.a,0,sizeof(c.a));22for(inti=1;i<=2;inc(i))23for(intj=1;j<=2;inc(j))24for(intk=1;k<=2;inc(k))25c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%P)%P;26returnc;27}28}A,X;29inlineJXKSM_2(JXa,intq,intP)30{//矩阵优化斐波拉契数列31JXAns=X,x=a;32while(q)33{34if(q&1)Ans=Ans*x;35q>>=1,x=x*x;36}37returnAns;38}39inlinevoidGet_Prime()40{//得出质数便于之后分解质因数41No_Prime[1]=1;42for(inti=2;i<=N;inc(i))43{44if(!No_Prime[i])Prime[inc(tot)]=i;45for(intj=1;j<=tot&&i*Prime[j]<=N;inc(j))46{47No_Prime[i*Prime[j]]=1;48if(i%Prime[j]==0)break;49}50}51}52inlinevoidResolve(intn,intPri[],intCnt[])53{//分解质因数,pri存质因子,cnt存次数54cnt=1;55intt=sqrt(n),TOT;56for(inti=1;Prime[i]<=t;inc(i))57if(n%Prime[i]==0)58{59TOT=0;60Pri[cnt]=Prime[i];61while(n%Prime[i]==0)62{63n/=Prime[i];64inc(TOT);65}66Cnt[cnt]=TOT;67inc(cnt);68}69if(n^1)70{71Pri[cnt]=n,Cnt[cnt]=1;72inc(cnt);73}74dec(cnt);75}7677inlineintKSM_1(intx,intq,intP)78{//普通的快速幂79intAns=1;80x%=P;81while(q)82{83if(q&1)Ans=Ans*x%P;84q>>=1,x=x*x%P;85}86returnAns;87}88inlinevoidWork(intn)89{//求n的因数90TOT=0;91intt=sqrt(n);92for(inti=1;i<=t;inc(i))93{94if(n%i==0)95{96if(i*i==n)Factor[inc(TOT)]=i;97else98{99Factor[inc(TOT)]=i;100Factor[inc(TOT)]=n/i;101}102}103}104}105inlineintGcd(inta,intb)106{107returna%b==0?b:Gcd(b,a%b);108}109inlineintFind_Loop(intn)110{111Resolve(n,Pri,Cnt);112intAns=1,loop;113for(inti=1;i<=cnt;inc(i))114{115loop=1;116P=Pri[i];117switch(P)118{119case2:loop=3;break;120case3:loop=8;break;121case5:loop=20;break;122default:123{124if(KSM_1(5,(P-1)>>1,P)==1)125Work(P-1);126else127Work(2*(P+1));128sort(Factor,Factor+TOT+1);129for(intj=1;j<=TOT;inc(j))130{131JXB=KSM_2(A,Factor[j]-1,P);132intx=(B.a[1][1]+B.a[1][2])%P;133inty=(B.a[2][1]+B.a[2][2])%P;134if(x==1&&y==0)135{136loop=Factor[j];137break;138}139}140}break;141}142loop*=KSM_1(P,Cnt[i]-1,9223372036854775807);143if(loop<Ans)swap(loop,Ans);144Ans=Ans/Gcd(loop,Ans)*loop;145}146returnAns;147}148inlineintMod(char*a,intb)//高精度a除以低精度b149{150intd=0,l=strlen(a);151for(inti=0;i<l;inc(i))d=(d*10+(a[i]-‘0‘))%b;//求出余数152returnd;153}154charpp[30000007];155signedmain()156{157A.a[1][1]=A.a[1][2]=A.a[2][1]=1;158X.a[1][1]=X.a[2][2]=1;159Get_Prime();160scanf("%s%lld",pp,&n);161intloop=Find_Loop(n);162intAns=Mod(pp,loop);163if(!Ans)Ans=loop;164if(Ans<=0)putchar(48);165else166if(Ans<=2)printf("%lld",1ll%n);167else168{169A.a[1][1]=A.a[1][2]=A.a[2][1]=1;170X.a[1][1]=X.a[2][2]=1;171P=n;172X=KSM_2(A,Ans-1,n);173printf("%lld",X.a[1][1]);174}175return0;176}
标签：$$ SM space 黄金分割 void struct 素数 signed pac
原文地址：https://www.cnblogs.com/Shine-Sky/p/9076792.html
