HDU 2256 Problem of Precision（矩阵快速幂）+ HDU 4565

HDU 2256

题意：

计算?(2√+3√)2n?mod1024

思路：

∵f(n)=(2√+3√)2n=(5+26√)n=An+Bn?6√

∴f(n?1)=An?1+Bn?1?6√

又∵f(n)=(5+26√)?f(n?1)

∴f(n)=(5?An?1+12?Bn?1)+(2?An?1+5?Bn?1)?6√

所以递推矩阵就是：

(52125)?(An?1Bn?1)=(AnBn)

A1=5,B1=2.

然后套矩阵快速幂模板即可求出An,Bn.

又∵(5+26√)n=An+Bn?6√

∴(5?26√)n=An?Bn?6√

∴(5+26√)n+(5?26√)n=2An

又∵(5?26√)n<1

∴?(5+26√)n?=2An?1

所以最后答案就是2An?1

代码：

/*
* @author FreeWifi_novicer
* language : C++/C
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>

using namespace std;

#define clr( x , y ) memset(x,y,sizeof(x))
#define cls( x ) memset(x,0,sizeof(x))
#define mp make_pair
#define pb push_back
typedef long long lint;
typedef long long ll;
typedef long long LL;

const int mod = 1024;
const int maxn = 4;
struct Matrix{
    int n,m;
    lint a[maxn][maxn];
    Matrix(int n , int m){
        this->n = n;
        this->m = m;
        cls(a);
    }
    Matrix operator * (const Matrix &tmp){
        Matrix res(n,tmp.m);
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < tmp.m ; j++){
                for(int k = 0 ; k < m ; k++){
                    res.a[i][j] = (res.a[i][j] + (a[i][k] * tmp.a[k][j]) % mod) % mod;
                }
            }
        }
        return res;
    }
};

void Matrix_print(Matrix x){
    for(int i = 0 ; i < x.n ; i++){
        for(int j = 0 ; j < x.m ; j++) cout << x.a[i][j] << ‘ ‘;
        cout << endl;
    }
}
Matrix fast_pow(Matrix x ,int n){
    Matrix res(x.n,x.m);
    for(int i = 0 ; i <  x.n ; i++) res.a[i][i] = 1;
    while(n){
        if(n&1)
            res = res * x;
        x = x*x;
        n >>= 1;
        //Matrix_print(res);
        //cout << endl;
    }
    return res;
}

void solve(){
    lint n;
    cin >> n;
    if(n == 1){
        cout << 9 << endl;
        return;
    }
    Matrix base(2,1);
    Matrix fun(2,2);
    base.a[0][0] = 5;
    base.a[1][0] = 2;
    fun.a[0][0] = 5;
    fun.a[0][1] = 12;
    fun.a[1][0] = 2;
    fun.a[1][1] = 5;
    fun = fast_pow(fun,n-1);
    base = fun * base;
    cout << (2*base.a[0][0] - 1) % mod << endl;
}

int main(){
    int t ; cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

附送HDU 4565代码:

/*
* @author FreeWifi_novicer
* language : C++/C
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>

using namespace std;

#define clr( x , y ) memset(x,y,sizeof(x))
#define cls( x ) memset(x,0,sizeof(x))
#define mp make_pair
#define pb push_back
typedef long long lint;
typedef long long ll;
typedef long long LL;
lint mod;
lint n , a , b ;
const int maxn = 4;
struct Matrix{
    int n,m;
    lint a[maxn][maxn];
    Matrix(int n , int m){
        this->n = n;
        this->m = m;
        cls(a);
    }
    Matrix operator * (const Matrix &tmp){
        Matrix res(n,tmp.m);
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < tmp.m ; j++){
                for(int k = 0 ; k < m ; k++){
                    res.a[i][j] = (res.a[i][j] + (a[i][k] * tmp.a[k][j]) % mod) % mod;
                }
            }
        }
        return res;
    }
};

void Matrix_print(Matrix x){
    for(int i = 0 ; i < x.n ; i++){
        for(int j = 0 ; j < x.m ; j++) cout << x.a[i][j] << ‘ ‘;
        cout << endl;
    }
}
Matrix fast_pow(Matrix x ,int n){
    Matrix res(x.n,x.m);
    for(int i = 0 ; i <  x.n ; i++) res.a[i][i] = 1;
    while(n){
        if(n&1)
            res = res * x;
        x = x*x;
        n >>= 1;
        //Matrix_print(res);
        //cout << endl;
    }
    return res;
}

void solve(){
    if(n == 1){
        lint ans = (lint)( a + ceil( sqrt( b * 1.0 ) ) ) % mod;
        cout << ans << endl;
        return;
    }
    Matrix base(2,1);
    Matrix fun(2,2);
    base.a[0][0] = a % mod;
    base.a[1][0] = 1;
    fun.a[0][0] = a % mod;
    fun.a[0][1] = b % mod;
    fun.a[1][0] = 1;
    fun.a[1][1] = a % mod;
    fun = fast_pow(fun,n-1);
    base = fun * base;
    cout << (2 * base.a[0][0]) % mod << endl;
}

int main(){
    while(cin >> a >> b >> n >> mod ){
        solve();
    }
    return 0;
}

版权声明：博主表示授权一切转载：）