快速幂,矩阵乘法,矩阵快速幂

Posted 2021-01-17 codemaker-li

快速幂利用二进制

复杂度 log级

#include <cstdio>
#include <iostream>
#include <string>
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;

int q_power(int a,int b,int c) {
    int r=1;
    a%=c;
    while (b) {
        if (b&1) {
            r=(r*a)%c;
        }
        a=(a*a)%c;
        b>>=1;
    }
    return r;
}

int a,b,c;
int main () {
    cin>>a>>b>>c;
    cout<<q_power(a,b,c);
    return 0;
}

附带上矩阵快速幂以及

矩阵快速幂求斐波那契数列：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int mod = 10000;
const int maxn = 35;
int N;
struct Matrix {
    int mat[maxn][maxn];
    int x, y;
    Matrix() {
        memset(mat, 0, sizeof(mat));
        for (int i = 1; i <= maxn - 5; i++) mat[i][i] = 1;
    }
};
inline void mat_mul(Matrix a, Matrix b, Matrix &c) {
    memset(c.mat, 0, sizeof(c.mat));
    c.x = a.x; c.y = b.y;
    for (int i = 1; i <= c.x; i++) {
        for (int j = 1; j <= c.y; j++) {
            for (int k = 1; k <= a.y; k++) {
                c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % mod;
                c.mat[i][j] %= mod;
            }
        }
    }
    return ;
}
inline void mat_pow(Matrix &a, int z) {
    Matrix ans, base = a;
    ans.x = a.x; ans.y = a.y;
    while (z) {
        if (z & 1 == 1) mat_mul(ans, base, ans);
        mat_mul(base, base, base);
        z >>= 1;
    }
    a = ans;
}
int main() {
    while (cin >> N) {
        switch (N) {
            case -1: return 0;
            case 0: cout << "0" << endl; continue;
            case 1: cout << "1" << endl; continue;
            case 2: cout << "1" << endl; continue;
        }
        Matrix A, B;
        A.x = 2; A.y = 2;
        A.mat[1][1] = 1; A.mat[1][2] = 1;
        A.mat[2][1] = 1; A.mat[2][2] = 0;
        B.x = 2; B.y = 1;
        B.mat[1][1] = 1; B.mat[2][1] = 1;
        mat_pow(A, N - 1);
        mat_mul(A, B, B);
        cout << B.mat[1][1] << endl;
    }
    return 0;
}

顺便来一发

矩阵乘法：

/*
    假设 A 是 m*p 的矩阵 , B 是 p*n 的矩阵
    记 C = AB (C 是 矩阵 A与B的乘积)
    那么 C 是 m*n 的矩阵 
*/
for (int i = 1;i <= m;++i)//A的行 
{
    for (int j = 1;j <= n;++j)//B的列 
    {
        for (int k = 1;k <= p;++k)//通过公式求C 
        {
            C[i][j] += A[i][k]*B[k][j];
        }
    }
}

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long LL;
const LL mod = 1000000007;
/*
    矩阵快速幂求斐波那契数列
    输入 n 输出 f[n] 
*/ 
struct Mat
{
    LL mat[2][2];
};
Mat operator * (Mat a,Mat b)//矩阵乘法
{
    Mat c;
    for (int i = 0;i < 2;++i)
    {
        for (int j = 0;j < 2;++j)
        {
            c.mat[i][j] = 0;
            for (int k = 0;k < 2;++k)
            {
                c.mat[i][j] = ((a.mat[i][k]*b.mat[k][j])%mod + c.mat[i][j])%mod;
            }
        }
    }
    return c;
}
Mat operator ^ (Mat a,LL k)//矩阵幂
{
    Mat c;
    for (int i = 0;i < 2;++i)
    {
        for (int j = 0;j < 2;++j)
        {
            c.mat[i][j] = (i==j);//初始化为单位矩阵
        }
    }
    //据说任何矩阵乘以单位矩阵其值不会变
 
    for (;k;k>>=1)
    {
        if (k&1) c = c*a;
        a = a*a;
    }
    return c;
}
int main()
{
    LL n;
    while (cin>>n)
    {
        Mat a;
        a.mat[0][0] = 1,a.mat[0][1] = 1,a.mat[1][0] = 1,a.mat[1][1] = 0;
        Mat fn = a^n;
        cout<<fn.mat[0][1]<<endl;
    }
    return 0;
}