快速幂及矩阵

基础快速幂

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
int qpow(int a,int b,int mod)
{
    int res=1;
    for(ll i=1;i<=b;i<<=1,a=((ll)a*a)%mod)
        if(i&b)res=((ll)res*a)%mod;
    return res;
}
int main()
{
    ll b, p, k;
    cin>>b>>p>>k;
    cout<<b<<"^"<<p<<" mod "<<k<<"="<<qpow(b,p,k);
    return 0;
}

矩阵快速幂

#include<iostream>
#include<cstdio>
using namespace std;
const int P=1e9+7; 
typedef long long ll;
struct matrix
{
    int s[2][2];
};
matrix mul(matrix a, matrix b)
{
    matrix ans;
    int i,j,k;
    for(i = 0; i < 2; i ++)
        for(j = 0; j < 2; j ++)
            for(k = 0; k < 2; k ++)
                ans.s[i][j]=((ll)ans.s[i][j] + (ll)(a.s[i][k] * b.s[k][j]) % P) % P;
    return ans;
}
matrix power(matrix a, int n)
{
    matrix im;//单位矩阵 
    im.s[0][0] = im.s[1][1] = 1;
    im.s[1][0] = im.s[0][1] = 0;
    matrix b = a;
    for(int i = 1; i <= n; i <<= 1, b = mul(b, b))
        if(n & i) im = mul(im, b);
    return im;
}

应用：求斐波那契数列

int F(int n)
{
    if (n == 0) return 1 % P;
    else if(n == 1) return 1 % P;
    else
    {
        matrix a;
        a.s[0][0] = a.s[0][1] = a.s[1][0] = 1;
        a.s[1][1] = 0;
        a = power(a, n - 2);
        int f = (long long) (a.s[0][0] + a.s[0][1]) % P;
        return f;
    }
}
int main()
{
    int n;
    scanf("%d", & n);
    cout << F(n) << endl;
    return 0;
}