HDU 5950 Recursive sequence 递推转矩阵

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2882    Accepted Submission(s): 1284

Problem 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 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.

 

 

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.

 

题意     f[n]=f[n-1]+2*f[n-2]+n^4; f[1]=a f[2]=b   求第n项

解析     直接递推肯定会超时的   所以 构造一个7*7系数矩阵 直接快速幂解出来  由于现在比较菜 只会最简单的矩阵 勉强可以写出来。。。。

 

1 2 1 0 0 0 0     　　  f[i-1]　　　　　　　　f[i]

1 0 0 0 0 0 0      　　 f[i-2]　　　　　　　　f[i-1]　　

0 0 1 4 6 4 1　　　　i^4　　　　　　　　  (i+1)^4

0 0 0 1 3 3 1       *   　i^3                  =            (i+1)^3

0 0 0 0 1 2 1　　　　i^2　　　　　　　　  (i+1)^2

0 0 0 0 0 1 1　　　　i　　　　　　　　　   i+1

0 0 0 0 0 0 1　　　   1　　　　　　　　　  1　　

 

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=2147493647,maxn=7;
struct Matrix
{
    ll m[maxn][maxn];
    Matrix()
    {
        memset(m,0,sizeof(m));
    }
    void init()
    {
        for(int i=0; i<maxn; i++)
            for(int j=0; j<maxn; j++)
                m[i][j]=(i==j);
    }
    Matrix  operator +(const Matrix &b)const
    {
        Matrix c;
        for(int i=0; i<maxn; i++)
        {
            for(int j=0; j<maxn; j++)
            {
                c.m[i][j]=(m[i][j]+b.m[i][j])%mod;
            }
        }
        return c;
    }
    Matrix  operator *(const Matrix &b)const
    {
        Matrix c;
        for(int i=0; i<maxn; i++)
        {
            for(int j=0; j<maxn; j++)
            {
                for(int k=0; k<maxn; k++)
                {
                    c.m[i][j]=(c.m[i][j]+(m[i][k]*b.m[k][j])%mod)%mod;
                }
            }
        }
        return c;
    }
    Matrix  operator^(const ll &t)const
    {
        Matrix ans,a=(*this);
        ans.init();
        ll n=t;
        while(n)
        {
            if(n&1) ans=ans*a;
            a=a*a;
            n>>=1;
        }
        return ans;
    }
};
int main()
{
    int t;
    ll n,m,a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %lld %lld",&n,&a,&b);
        if(n==1)
        {
            cout<<a%mod<<endl;
            continue;
        }
        if(n==2)
        {
            cout<<b%mod<<endl;
            continue;
        }
        Matrix temp;
        temp.m[0][0]=1,temp.m[0][1]=2,temp.m[0][2]=1,temp.m[0][3]=0,temp.m[0][4]=0,temp.m[0][5]=0;temp.m[0][6]=0;
        temp.m[1][0]=1,temp.m[1][1]=0,temp.m[1][2]=0,temp.m[1][3]=0,temp.m[1][4]=0,temp.m[1][5]=0;temp.m[1][6]=0;
        temp.m[2][0]=0,temp.m[2][1]=0,temp.m[2][2]=1,temp.m[2][3]=4,temp.m[2][4]=6,temp.m[2][5]=4;temp.m[2][6]=1;
        temp.m[3][0]=0,temp.m[3][1]=0,temp.m[3][2]=0,temp.m[3][3]=1,temp.m[3][4]=3,temp.m[3][5]=3;temp.m[3][6]=1;
        temp.m[4][0]=0,temp.m[4][1]=0,temp.m[4][2]=0,temp.m[4][3]=0,temp.m[4][4]=1,temp.m[4][5]=2;temp.m[4][6]=1;
        temp.m[5][0]=0,temp.m[5][1]=0,temp.m[5][2]=0,temp.m[5][3]=0,temp.m[5][4]=0,temp.m[5][5]=1;temp.m[5][6]=1;
        temp.m[6][0]=0,temp.m[6][1]=0,temp.m[6][2]=0,temp.m[6][3]=0,temp.m[6][4]=0,temp.m[6][5]=0;temp.m[6][6]=1;
        Matrix aa,bb;
        bb.m[0][0]=b%mod;
        bb.m[1][0]=a%mod;
        bb.m[2][0]=81;
        bb.m[3][0]=27;
        bb.m[4][0]=9;
        bb.m[5][0]=3;
        bb.m[6][0]=1;
        aa=temp^(n-2);
        aa=aa*bb;
        cout<<aa.m[0][0]%mod<<endl;
    }
    return 0;
}