hdu-5667 Sequence (Matrix fast Power + Fermat theorem + fast Power)

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Sequence

Time limit:2000/1000 MS (java/others)

Memory limit:65536/65536 K (java/others)

problem Descriptionholion August would eat every thing he has found.
Now there is many foods,but he does not want to eat all of the them at Once,so he find a sequence.

FN=?????1,AB,ABFCN−1FN−2,N=1n=2otherWIse

He gives you 5 numbers N,a,b,c,p,and he'll eat fn Foods. But there is only p foods,so your should tell him fn mod p.

InputThe first line has a Number,t,means testcase.

Each testcase have 5 numbers,including n,a,b,c,p in a line.

1≤T≤10,1≤n≤10^,1≤a,b,c≤10^9 , P is a prime number,and p≤10^9+7.

Output Output one number for each Case,which is fn mod p.

Sample Input 15 3 3 3 233

Sample Output 190 Test Instructions:Ask F (n) The result of P modulo; Ideas:(ab) p[n]= AB * ((AB) p[n-1]) c * ((AB) p[n-2]); Recursive formulas can be written like this, merged into (AB) p[n]= (AB) (c*p[n-1]+p[n-2]+1); p[n]=c*p[n-1]+p[n-2]+ can be obtained 1; Such recursion can be obtained by matrix multiplication of the nth item; This is an application of matrix multiplication, to the matrix67 great God's blog address can learn, dot here to construct matrix multiplication: P[n] C 1 1 p[n-1]p[n-1]= 1 0 0 * p[n-2]1 0 0 1 1 Then there is a problem in the middle, that is, the problem of modeling;ab*p[n]%mod=ab*p[n]% (mod-1)%mod;This is obtained according to the Fermat theorem; a (p-1) ξ1%p; here's the address.ab*p[n]%mod=ab*p[n]/(mod-1) * (mod-1) +b*p[n]% (mod-1)%mod;make x=b*p[n]/(mod-1) ab*p[n]%mod=ax* (mod-1) +b*p[n]% (mod-1 )%mod=ab*p[n]% (mod-1)%mod;This is the result of taking the mold;and then the answer is fast power calculation;also have to pay attention to is the situation of a%p==0; AC Code:

/*5667 0MS 1584K 1835B g++ 2014300227*/#include<bits/stdc++.h>using namespacestd;Const intn=1e4+5; typedefLong Longll;Const intmod=1e9+7; ll N,a,b,c,p;structmatrix{ll a[3][3];}; Matrix A;voidIint (ll x)//matrix initialization; {a.a[0][0]=x; a.a[0][1]=1; a.a[0][2]=1; a.a[1][0]=1; a.a[1][1]=a.a[1][2]=0; a.a[2][0]=a.a[2][1]=0; a.a[2][2]=1;}    Matrix Mul (Matrix X,matrix y)//matrices multiplied {matrix ans;  for(intI=0;i<3; i++)    {         for(intj=0;j<3; j + +) {Ans.a[i][j]=0;  for(intk=0;k<3; k++) {Ans.a[i][j]+ = (X.a[i][k]*y.a[k][j])% (P-1); ANS.A[I][J]%= (P-1); }        }    }    returnans;} ll fast_fun (Matrix Temp,ll num)//matrix fast power; {Matrix S,Base;  for(intI=0;i<3; i++)//s initialized to the unit matrix { for(intj=0;j<3; j + +) {S.a[i][j]=0; Base. a[i][j]=Temp.a[i][j]; }} s.a[0][0]=s.a[1][1]=s.a[2][2]=1;  while(num) {if(num&1) {s=mul (s),Base); }        Base=mul (Base,Base); Num= (num>>1); }    return(s.a[0][0]+s.a[0][2])% (P-1);} ll Fastpow (ll fx,ll FY)//fast power to find results; {ll S=1,Base=FX;  while(FY) {if(fy&1) {s*=Base; S%=p; }        Base*=Base; Base%=p; FY= (fy>>1); }    returns;}intMain () {intT; scanf ("%d",&t);  while(t--) {scanf ("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);        Iint (c); if(n==1) printf ("1\n"); Else if(n==2) printf ("%lld\n", Fastpow (A, b)); Else        {            if(a%p==0) printf ("0\n"); Else{ll GG=fast_fun (a,n-2) *b% (P-1); printf ("%lld\n", Fastpow (A,GG)); }        }    }    return 0;}

hdu-5667 Sequence (Matrix fast Power + Fermat theorem + fast Power)