hdu 5950 Recursive sequence 矩阵快速幂

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

　　套板子，取模开ll就好了；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,MOD=;
struct Matrix
{
    ll a[][];
    Matrix()
    {
        memset(a,,sizeof(a));
    }
    void init()
    {
        for(int i=;i<;i++)
            for(int j=;j<;j++)
                a[i][j]=(i==j);
    }
    Matrix operator + (const Matrix &B)const
    {
        Matrix C;
        for(int i=;i<;i++)
            for(int j=;j<;j++)
                C.a[i][j]=(a[i][j]+B.a[i][j])%MOD;
        return C;
    }
    Matrix operator * (const Matrix &B)const
    {
        Matrix C;
        for(int i=;i<;i++)
            for(int k=;k<;k++)
                for(int j=;j<;j++)
                    C.a[i][j]=(C.a[i][j]+(a[i][k]*B.a[k][j])%MOD)%MOD;
        return C;
    }
    Matrix operator ^ (const ll &t)const
    {
        Matrix A=(*this),res;
        res.init();
        int p=t;
        while(p)
        {
            if(p&)res=res*A;
            A=A*A;
            p>>=;
        }
        return res;
    }
};
Matrix base,hh;
void init()
{
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
    base.a[][]=;
}
void init1(ll a,ll b)
{
    memset(hh.a,,sizeof(hh.a));
    hh.a[][]=b%MOD;
    hh.a[][]=a%MOD;
    hh.a[][]=***;
    hh.a[][]=**;
    hh.a[][]=*;
    hh.a[][]=;
    hh.a[][]=;
}
int main()
{
    init();
    int T,cas=;
    scanf("%d",&T);
    while(T--)
    {
        ll n,a,b;
        scanf("%lld%lld%lld",&n,&a,&b);
        init1(a,b);
        Matrix ans=(base^(n-));
        hh=hh*ans;
        printf("%lld\n",hh.a[][]);
    }
    return ;
}