标签：matrix int ll 单位矩阵 maxn ans 快速

快速幂

 例题：P1226 【模板】快速幂||取余运算

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int a, b, p;
int main() {
    cin >> a >> b >> p;
    int aa = a, bb = b;
    ll ans = 1;
    while (b) {
        if (b % 2 == 1) {
            ans = (ll)ans * a % p;
        }
        b /= 2;
        a = (ll) a * a % p;
    }
    printf("%d^%d mod %d=%ld\n", aa, bb, p, ans);

    return 0;
}

 

 

矩阵快速幂

　1.矩阵乘法

for(i = 1; i <= r; i ++) {
    for(j = 1; j <= c; j ++) {
        mul[i][j] = 0; 
        for(k = 1; k <= c; k ++) { 
            mul[i][j] += a[i][k] * b[k][j]; 
        } 
    } 
} 

 

　2.重载运算符

matrix operator *(const matrix &x, const matrix &y) {
     matrix z;
     for(int i = 1; i <= n; i ++) {
         for(int j = 1; j <= n; j ++) {
             for(int k = 1; k <= n; k ++) {
                 z.a[i][j] = (z.a[i][j] + x.a[i][k] * y.a[k][j] % mod) % mod;             
             }
         }
     }     
     return z;
}

 

　3.单位矩阵

　　误区：单位矩阵是左上右下对角线上元素全为1，并不是所有元素全为1。

struct matrix {
    ll a[maxn][maxn];
    matrix() {
        memset(a, 0, sizeof(a));
    }
    //构造单位矩阵 
    inline void build() {
        for(int i = 1; i <= n; i ++) {
            a[i][i] = 1; 
        }
    }
};

 

  P3390 【模板】矩阵快速幂  

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 150;
const int mod = 1e9 + 7;
int n;
ll k;

struct matrix {
    ll a[maxn][maxn];
    matrix() {
        memset(a, 0, sizeof(a));
    }
    //构造单位矩阵 
    inline void build() {
        for(int i = 1; i <= n; i ++) {
            a[i][i] = 1; 
        }
    }
}m;

//重载运算符
matrix operator *(const matrix &x, const matrix &y) {
     matrix z;
     for(int i = 1; i <= n; i ++) {
         for(int j = 1; j <= n; j ++) {
             for(int k = 1; k <= n; k ++) {
                 z.a[i][j] = (z.a[i][j] + x.a[i][k] * y.a[k][j] % mod) % mod;             
             }
         }
     }
             
     return z;
}

int main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= n; j ++) {
            cin >> m.a[i][j];
        }
    }
    matrix ans;
    ans.build();
    
    while(k) {
        if(k & 1) {//奇数（不能被2整除） 
            ans = ans * m;
        }
        k >>= 1;
        m = m * m;
    }
    
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= n; j ++) {
            cout << ans.a[i][j] << " ";
        }
        cout << "\n";
    }
    
    return 0;
}

 

标签：matrix,int,ll,单位矩阵,maxn,ans,快速
From： https://www.cnblogs.com/coding-inspirations/p/16621506.html