HDU 5674 Function(斐波那契模数列循环节)

Problem Description
There is a function
f(n)=(a+√b)^n+(a−√b)^n. a and b are integers
(1≤a,b≤1,000,000).
Maybe the function looks complex but it is actually an integer.The question is to calculate
f(x^y).The answer can bevery large,so just output the answer mod 1,000,000,007.

Input
There are multiple test cases. The first line of input contains an integer
T(1≤T≤200)
indicating the number of test cases. For each test case:

One line contains four integers
a,b (1≤a,b≤1,000,000), x(1≤x≤50),y(1≤y≤10^18).

Output
For each test case, output one integer.

Sample Input
3
3 5 1 1
3 5 2 1
1 1 49 99999

Sample Output
6
28
160106184

斐波那契数列循环节的定理就不在这里证明了,结论就是:假设MOD为模数,则在f[2]!=MOD时,(MOD+1)(MOD-1)一定为其一个循环节.这道题f[n]明显是裴波那契数列通项的形式,所以可以利用裴波那契数列的特征方程来构造裴波那契数列 f[n]=2a*f[n-1]+(b-a^2)*f[n-2]; 之后就可以利用循环节+快速幂求出答案了.

#include <stdio.h>#include <iostream>#include <algorithm>#include <string>#include <string.h>using namespace std;#define ll long long #define F(x,a,b) for (int x=a;x<=b;x++)#define _fast F(i,1,2)F(j,1,2)F(k,1,2) b[i][j]=((b[i][j]+(a[i][k]%MOD)*(a[k][j]%MOD))%MOD+MOD)%MOD;#define _reset F(i,1,2)F(j,1,2) {a[i][j]=b[i][j];b[i][j]=0;}#define _c F(i,1,2)F(j,1,2)F(k,1,2) b[i][j]=((b[i][j]+(a[i][k]%MOD)*(c[k][j]%MOD))%MOD+MOD)%MOD;ll a[4][4],b[4][4],c[4][4];ll f[3];const ll MOD = 1e9+7;void Init(ll aa,ll bb){    a[1][1]=aa;a[1][2]=bb;    a[2][1]=1;a[2][2]=0;    for (int i=1;i<=2;i++)        for (int j=1;j<=2;j++) c[i][j]=a[i][j];}ll multi(ll x,ll y,ll mod){    ll ans=0;    while (y)    {        if (y&1) ans=(ans+x+mod)%mod;        x=((x<<1)+mod)%mod;        y=y>>1;    }    return ans;}void fastmat(ll x){    if (x<=1) return;    if (x&1) {fastmat(x-1);_c _reset }    else     {        fastmat(x/2); _fast _reset    }}ll _f(ll x,ll k, ll mod){    if (!k) return 1;    if (k&1) return multi(_f(x,k-1,mod),x,mod);    else    {        ll t=_f(x,k/2,mod);        return multi(t,t,mod);    }}void out(){    for (int i=1;i<=2;i++)    {        for (int j=1;j<=2;j++)            printf("%I64d ",a[i][j]);        printf("\n");    }}int main(){     int T;     scanf("%d",&T);     while (T--)     {         ll aa,bb,x,y;         scanf("%I64d%I64d%I64d%I64d",&aa,&bb,&x,&y);         f[1]=aa*2;f[2]=(aa*aa+bb)*2;         ll a1=2*aa,b1=bb-aa*aa;         Init(a1,b1);         if (b1==0)          {            ll tt=_f(x,y,MOD-1);            printf("%I64d\n",_f(f[1],tt,MOD));         }else          {            ll mmod=(MOD-1)*(MOD+1);            ll tt=_f(x,y,mmod);            if (!tt)            {                fastmat(mmod-2);                printf("%I64d\n",(a[1][1]%MOD*f[2]%MOD+a[1][2]%MOD*f[1]%MOD)%MOD);            }else             {                if (tt<=2){printf("%I64d\n",f[tt]%MOD);}                else                 {                  fastmat(tt-2);                  printf("%I64d\n",((a[1][1]%MOD)*(f[2]%MOD)+(a[1][2]%MOD)*(f[1]%MOD))%MOD);                  //out();                  }            }         }     }}