HDU 4549 M斐波那契数列

首先找到f[n]，指数可以通过矩阵快速幂来求解，由于指数较大，需要利用以下公式降幂

得到指数之后，再利用快速幂得到答案。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const long long mod1=1000000006;
const long long mod2=1000000007;
long long a,b;
int n;
long long z1,z2;

long long mod(long long a, long long b)
{
    if (a >= 0) return a%b;
    if (abs(a) % b == 0) return 0;
    return (a + b*(abs(a) / b + 1));
}

struct Matrix
{
    long long A[5][5];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = mod((c.A[i][j] + mod(A[i][k] * b.A[k][j], mod1)), mod1);
    c.R=R; c.C=b.C;
    return c;
}

void init()
{
    Z.A[1][1] = 0, Z.A[1][2] = 1; Z.R = 1; Z.C = 2;
    Y.A[1][1] = 1, Y.A[1][2] = 0, Y.A[2][1] = 0, Y.A[2][2] = 1; Y.R = 2; Y.C = 2;
    X.A[1][1] = 0, X.A[1][2] = 1, X.A[2][1] = 1, X.A[2][2] = 1; X.R = 2; X.C = 2;
}

void work1(int x)
{
    while (x)
    {
        if (x % 2 == 1) Y = Y*X;
        x = x >> 1;
        X = X*X;
    }
    Z = Z*Y;
    z2=mod(Z.A[1][1], mod1);
  //  printf("%lld
", mod(Z.A[1][1], mod1));
}

void work2(int x)
{
    while (x)
    {
        if (x % 2 == 1) Y = Y*X;
        x = x >> 1;
        X = X*X;
    }
    Z = Z*Y;
    z1=mod(Z.A[1][1], mod1);
   // printf("%lld
", mod(Z.A[1][1], mod1));
}

long long quickmod(long long a,long long b,long long m)
{
    long long ans = 1;
    while(b)
    {
        if(b&1)
        {
            ans = (ans*a)%m;
            b--;
        }
        b/=2;
        a = a*a%m;
    }
    return ans;
}

int main()
{
    while(~scanf("%lld%lld%d",&a,&b,&n))
    {
        if(n==0) printf("%lld
",a%mod2);
        else if(n==1) printf("%lld
",b%mod2);
        else
        {
            init();
            work1(n);
            init();
            work2(n-1);
          printf("%lld
",(quickmod(a,z1,mod2)*quickmod(b,z2,mod2))%mod2);
        }
    }
    return 0;
}