HDU 5667 Sequence Matrix fast Power + Fermat theorem

Sequence

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)

Problem Description    Holion 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.

Input    The 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≤1≤a,b,c≤9 , P is a prime number,and p≤9+7.

OutputOutput one number for each Case,which is F n mod p. c8>

Sample Input15 3 3 3 233

Sample Output190

Sourcebestcoder Round #80思路: is f (n) =f (n-1) *c+f (n-2) +b; A^f (n)%p=a^ (f (n)% (p-1))%p;//Pony Small Matrix Construction Blog: http://www.cnblogs.com/frog112111/archive/2013/05/19/3087648.html There's a pit point A. %p==0

#include <iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<Set>#include<map>using namespacestd;#definell __int64#defineMoD 1000000007#defineINF 999999999//#pragma COMMENT (linker, "/stack:102400000,102400000")ll A,b,c,n,p;struct  is{ll a[Ten][Ten];}; isJuzhenmul ( isA isB,ll Hang, ll lie) {    inti,t,j;  isans; memset (ANS.A,0,sizeof(ANS.A));  for(i=1; i<=hang;i++)     for(t=1; t<=lie;t++)     for(j=1; j<=lie;j++) {Ans.a[i][t]+ = (a.a[i][j]*b.a[j][t]); Ans.a[i][t]%= (P-1); }    returnans;} isQuickpow ( isAns isa,ll x) {     while(x) {if(x&1) Ans=juzhenmul (Ans,a,3,3); A=juzhenmul (A,a,3,3); X>>=1; }    returnans;} ll Quickpow1 (ll a,ll N) {ll mul=1; if(a%p==0)    return 0;  while(n) {if(n&1) mul*=a,mul%=p; N>>=1; A=a*A; A%=p; }    returnMul;}intMain () { isAnsBase, GG; intGGGG; scanf ("%d",&GGGG);  while(gggg--) {ll fn; scanf ("%i64d%i64d%i64d%i64d%i64d",&n,&a,&b,&c,&p); if(n>2) {memset (ANS.A,0,sizeof(ANS.A)); ans.a[1][1]=1; ans.a[2][2]=1; ans.a[3][3]=1; memset (BaseA0,sizeof(Base. a)); Base. a[1][1]=C; Base. a[1][2]=1; Base. a[2][1]=1; Base. a[3][1]=1; Base. a[3][3]=1; Ans=quickpow (ans,Base, N-2); gg.a[1][1]=b; gg.a[1][2]=0; gg.a[1][3]=b; Ans=juzhenmul (Gg,ans,1,3); FN=ans.a[1][1]; }        Else        {            if(n==1) FN=0; Elsefn=b; } printf ("%i64d\n", Quickpow1 (A,FN)); }    return 0;}

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HDU 5667 Sequence Matrix fast Power + Fermat theorem