HDU2971 Tower 【矩阵快速幂】

Tower

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2862    Accepted Submission(s): 686

Problem Description

Alan loves to construct the towers of building bricks. His towers consist of many cuboids with square base. All cuboids have the same height h = 1. Alan puts the consecutive cuboids one over another:

Recently in math class, the concept of volume was introduced to Alan. Consequently, he wants to compute the volume of his tower now. The lengths of cuboids bases (from top to bottom) are constructed by Alan in the following way:

1. Length a1 of the first square is one.

2. Next, Alan fixes the length a2 of the second square.

3. Next, Alan calculates the length an (n > 2) by 2*a2*(an-1)-(an-2). Do not ask why he chose such

a formula; let us just say that he is a really peculiar young fellow. For example, if Alan fixes a2 = 2, then a3 = 8 -a1 = 7; see Figure 1. If Alan fixes a2 = 1, then an = 1 holds for all n belong to N; see Figure 2.

Now Alan wonders if he can calculate the volume of tower of N consecutive building bricks. Help Alan and write the program that computes this volume. Since it can be quite large, it is enough to compute the answer modulo given natural number m.

 

Input

The input contains several test cases. The first line contains the number t (t <= 10^5) denoting the number of test cases. Then t test cases follow. Each of them is given in a separate line containing three integers a2,N,m (1 <= a2,m <= 10^9, 2 <= N <= 10^9) separated by a single space, where a2 denotes the fixed length of second square in step 2, while N denotes the number of bricks constructed by Alan.

 

Output

For each test case (a2,N,m) compute the volume of tower of N consecutive bricks constructed by Alan according to steps (1-3) and output its remainder modulo m.

 

Sample Input

  

   3
2 3 100
1 4 1000
3 3 1000000000
  

 

Sample Output

  

   54
4
299

   

    

     Hint
    

    
     
    

   
    
  

 

Source

Central European Programming Contest 2008

 

卡常....各种TLE

减少取模次数才能过

令p=2*a2;

a[n] = p * a[n-1] - a[n-2]

①:a[n]^2 = (p * a[n-1])^2 + a[n-2]^2 - 2*p*a[n-1]*a[n-2]

②:v[n] = v[n-1] + a[n]^2

2*p*a[n]*a[n-1] = 2*p*( p*a[n-1] - a[n-2] ) * a[n-1]

③:2*p*a[n]*a[n-1] = 2*(p*a[n-1])^2 - 2*p*a[n-1]*a[n-2]

于是构造一个4*4矩阵即可

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<string.h>
#include<math.h>
#include<list>

using namespace std;

#define ll long long
#define pii pair<int,int>

const int inf=1e9+7;
const int INF=inf;
const double EPS=1e-8;

ll MOD=inf;

struct Matrix{
    const static int MAXN=4;
    ll m[MAXN][MAXN];

    void init0(){
        for(int i=0;i<MAXN;++i){
            fill(m[i],m[i]+MAXN,0);
        }
    }

    void initE(){//初始化为单位矩阵
        for(int i=0;i<MAXN;++i){
            for(int j=0;j<MAXN;++j){
                m[i][j]=(i==j);
            }
        }
    }

    ll* operator[](std::size_t n){
        return m[n];
    }

    const ll* operator[](std::size_t n)const{
        return m[n];
    }

    Matrix operator*(const Matrix&x){
        Matrix ans;
        for(int i=0;i<MAXN;++i){
            for(int j=0;j<MAXN;++j){
                ans[i][j]=0;
                for(int k=0;k<MAXN;++k){
                    ans[i][j]=(ans[i][j]+m[i][k]*x[k][j])%MOD;
                }
            }
        }
        return ans;
    }

    Matrix operator^(ll n){
        Matrix ans;
        ans.initE();
        Matrix t=*this;
        while(n){
            if(n&1){
                ans=ans*t;
            }
            t=t*t;
            n>>=1;
        }
        return ans;
    }
}mat1,mat2;

void init(ll a2){
    ll p=(2*a2)%MOD;
    mat1.init0();
    mat2.init0();

    mat2[0][0]=mat2[0][2]=mat2[1][2]=mat2[2][1]=1;
    mat2[0][1]=mat2[1][1]=(p*p)%MOD;
    mat2[3][1]=(2*p*p)%MOD;
    mat2[0][3]=mat2[1][3]=mat2[3][3]=-1+MOD;

    mat1[0][0]=(1+a2*a2)%MOD;
    mat1[1][0]=(a2*a2)%MOD;
    mat1[2][0]=1;
    mat1[3][0]=(2*p*a2)%MOD;
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        ll a2,n;
        scanf("%lld%lld%lld",&a2,&n,&MOD);

        init(a2);
        mat2=mat2^(n-2);

        int ans=(mat2*mat1)[0][0];
        ans=(ans%MOD+MOD)%MOD;
        printf("%d\n",ans);
    }
    return 0;
}