矩阵快速幂 模板

Posted 2021-01-10 jk17211764

/*
矩阵快速幂模板   斐波那契
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 2;
const int mod = 10000;

//矩阵结构体
struct Matrix{
    int a[maxn][maxn];
    void init(){    //初始化为单位矩阵
        memset(a, 0, sizeof(a));
        for(int i=0;i<maxn;++i){
            a[i][i] = 1;
        }
    }
};

//矩阵乘法
Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i=0;i<maxn;++i){
        for(int j=0;j<maxn;++j){
            ans.a[i][j] = 0;
            for(int k=0;k<maxn;++k){
                ans.a[i][j] += a.a[i][k] * b.a[k][j];
                ans.a[i][j] %= mod;
            }
        }
    }
    return ans;
}

//矩阵快速幂
Matrix qpow(Matrix a, int n){
    Matrix ans;
    ans.init();
    while(n){
        if(n&1) ans = mul(ans, a);
        a = mul(a, a);
        n /= 2;
    }
    return ans;
}

void output(Matrix a){
    for(int i=0;i<maxn;++i){
        for(int j=0;j<maxn;++j){
            cout << a.a[i][j] << " ";
        }
        cout << endl;
    }
}

int main(){
    Matrix a;
    a.a[0][0] = 1;
    a.a[0][1] = 1;
    a.a[1][0] = 1;
    a.a[1][1] = 0;

    Matrix ans = qpow(a, 10);

    cout << "f(12) = [f(2), f(1)] = [1, 1] * a^10 = " << ans.a[0][0] + ans.a[1][0];

    return 0;
}

 

                                   Recursive sequence

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 HDU 5950

Description

     Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and ${\mathrm{i4}}^{}$. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < ${231}^{}$

as described above.

 

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const long long mod = 2147493647;
const int MAXN = 8;
struct prog
{
    long long a[MAXN][MAXN];
};
prog s,B;
prog matrixmul(prog a,prog b)  //矩阵相乘
{
    prog c;
    for(int i=1; i<MAXN; ++i)
        for(int j=1; j<MAXN; ++j)
        {
            c.a[i][j]=0;
            for(int k=1; k<MAXN; k++)
                c.a[i][j]+=(a.a[i][k]*b.a[k][j])%mod;
            c.a[i][j]%=mod;
        }
    return c;
}
prog mul(prog s,int k)
{
    prog ans;
    for(int i=1; i<MAXN; ++i)
        for(int j=1; j<MAXN; ++j)
          ans.a[i][j]=(i==j)?1:0;
    while(k)
    {
        if(k&1)
            ans=matrixmul(ans,s);
        k>>=1;
        s=matrixmul(s,s);
    }
    return ans;
}
int main()
{
    int n,t,a,b;
    for(scanf("%d",&t); t--;)
    {
        scanf("%d %d %d",&n,&a,&b);
        if(n==1)
        {
            printf("%lld
",a%mod);
            continue;
        }
        if(n==2)
        {
            printf("%lld
",b%mod);
            continue;
        }
        if(n==3)
        {
            printf("%lld
",(81+2*a%mod+b%mod)%mod);
            continue;
        }
        n-=2;
        for(int i=1; i<=7; ++i)
            for(int j=1; j<=7; ++j)
              s.a[i][j]=0,
              B.a[i][j]=0;
        for(int i=1; i<=5; i++)
             s.a[i][1]=1;
        for(int i=2; i<=5; i++)
             s.a[i][2]=i-1;
        s.a[3][3]=1;
        s.a[4][3]=3;
        s.a[5][3]=6;
        s.a[4][4]=1;
        s.a[5][4]=4;
        s.a[5][5]=1;
        s.a[6][5]=1;
        s.a[6][6]=1;
        s.a[7][6]=1;
        s.a[6][7]=2;
        B.a[1][1]=1;
        B.a[2][1]=3;
        B.a[3][1]=9;
        B.a[4][1]=27;
        B.a[5][1]=81;
        B.a[6][1]=b;
        B.a[7][1]=a;
        s=mul(s,n);
        s=matrixmul(s,B);
        printf("%lld
",s.a[6][1]%mod);

        


    }
    return 0;
}

View Code

 

Sample Input

2 3 1 2 4 1 10

 

Sample Output

85 369

Hint

In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.