Hdu 5895 mathematician QSC exponential cyclic section + Matrix fast Power

Mathematician QSC

Time limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)

Problem DESCRIPTIONQSC dream of becoming a mathematician, he believes that everything in this world have a mathematical law .

Through unremitting efforts, one day he finally found the QSC sequence, it's a very magical sequence, can be calculated B Y a series of calculations to predict the results of a course of a semester of a student.

This sequence was such like that, first of all,F(0)=0,F(1)=1,F(N)=f (n− 2) +2∗ f ( N−1 ) (n ≥2 ) /span> Then the definition of the QSC sequence is  g ( N) =∑n i =0 F (i) 2 . If We know the birthday of the student is N, the year at the beginning of the semester are Y, the course number x and the C Ourse total score S, then the forecast Mark is xg(n∗y)%(s+1) .
QSC sequence published caused a sensation, after a number of students to find out the results of the prediction are very AC Curate, the shortcoming is the complex calculation. As clever as is, can you write a program to predict the mark?

Inputfirst line was an integer T (1≤t≤1000).

The next T lines were given n, y, X, S, respectively.

N, x is 8 bits decimal integer, for example, 00001234.

Y is 4 bits decimal integer, for example, 1234.
n, x, Y is not negetive.

1≤s≤100000000

Outputfor Each test case the output was only one integer number ans in a line.

Sample Input220160830 2016 12345678 66620101010 2014 03030303 333

Sample Output1317

Source2016 ACM/ICPC Asia Regional Shenyang Online

Idea: First of all, the a^b%c=a^ (B%phi (c) +phi (c))%c B>=phi (c) Index follow-up link;

Then, the G function, f (n) can obviously be used to quickly write a matrix, g (n) =f (n) *f (n+1)/2; because/2, modulo division, the first thought of the inverse, but the mold is not necessarily odd, even the case 2 without inverse element;

Now how to deal with the 2,F (n) and F (n+1) must have an even number, found that the recurrence after 2 is changed to f (n) =6*f (n-1)-F (n-2);

PS: A small trick to deal with 2,MDZZ, modulo the number of * *, answer/2;

See Code;

#include <bits/stdc++.h>using namespacestd;#definell Long Long#definePi (4*atan (1.0))Const intn=1e5+Ten, m=1e6+1010, inf=1e9+Ten, mod=1e9+7;Constll inf=1e18+Ten; ll N,x,y,s;ll m;struct  is{ll a[Ten][Ten];}; isJuzhenmul ( isA isB,ll Hang, ll lie,ll MoD) {    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]%=MoD; }    returnans;} isQuickpow ( isAns isa,ll x,ll MoD) {     while(x) {if(x&1) Ans=juzhenmul (Ans,a,2,2, MoD); A=juzhenmul (A,a,2,2, MoD); X>>=1; }    returnans;}voidExtend_euclid (ll A, ll B, ll &x, LL &y) {    if(b = =0) {x=1; Y=0; return; } extend_euclid (b, a%b, x, y); LL TMP=x; X=y; Y= tmp-(A/b) *y;} ll Phi (ll N) {ll I,rea=N;  for(i=2; i*i<=n;i++)    {        if(n%i==0) {rea=rea-rea/i;  while(n%i==0) n/=i; }    }    if(n>1) Rea=rea-rea/N; returnRea;} ll Pow (ll a,ll n,ll MoD) {ll ans=1;  while(n) {if(n&1) {ans=ans*a%MoD; } A=a*a%MoD; N>>=1; }    if(ans==0) ans+=MoD; returnans;} ll Getans (ll x,ll MoD) {if(x==0)    return 0; if(x==1)    return 1;  isAnsBase; memset (ANS.A,0,sizeof(ANS.A)); ans.a[1][1]=1; ans.a[2][2]=1; Base. a[1][1]=0; Base. a[1][2]=1; Base. a[2][1]=1; Base. a[2][2]=2; Ans=quickpow (ans,Base, X-2, MoD); return(ans.a[2][1]+ans.a[2][2]*2)%MoD;} ll Getans2 (ll x,ll MoD) {if(x==0)    return 0; if(x==1)    return 1;  isAnsBase; memset (ANS.A,0,sizeof(ANS.A)); ans.a[1][1]=1; ans.a[2][2]=1; Base. a[1][1]=6; Base. a[1][2]=-1; Base. a[2][1]=1; Base. a[2][2]=0; Ans=quickpow (ans,Base, X-2, MoD); return((((ans.a[1][1]*6-ans.a[2][1])%mod) +mod)%mod);}intMain () {intT; scanf ("%d",&T);  while(t--) {scanf ("%lld%lld%lld%lld",&n,&y,&x,&s); ll Zhi=n*y; M=phi (s+1);        ll K; if(zhi%2==0) K= (Getans (zhi+1, m)%m) * (Getans2 (zhi/2, m)%m)%m; Elsek= (Getans (zhi,m)%m) * (Getans2 (zhi+1)/2, m)%m)%m; ll out=pow (x,k,s+1); printf ("%lld\n", out); }    return 0;}

Hdu 5895 mathematician QSC exponential cyclic section + Matrix fast Power