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)
Total Submission(s): 1323    Accepted Submission(s): 589

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）

 

Recommend

jiangzijing2015

题解：

　　转移方程 Fn = 2*Fn-2 + Fn-1 + n^4。如果没有这个n^4，这题就会很简单，所以我们就需要化简n^4。

　　i^4 = C(0,4) (i-1)^4 + C(1,4)*(i-1)^3 + C(2,4)*(i-1)^2 + C(3,4)*(i-1)^1 + C(4,4)*(i-1)^0

　　所以 n^4 = (n-1)^4 + 4*(n-1)^3 + 6*(n-1)^2 + 4*(i-1) + 1。

　　Fn = 2*Fn-2 + Fn-1 + (n-1)^4 + 4*(n-1)^3 + 6*(n-1)^2 + 4*(i-1) + 1。

 

　　

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <set>
using namespace std;
typedef long long LL;
#define ms(a, b) memset(a, b, sizeof(a))
#define pb push_back
#define mp make_pair
const int INF = 0x7fffffff;
const int inf = 0x3f3f3f3f;
const LL mod = 2147493647;
const int maxn = 500+10;
struct Matrix
{
    LL c[7][7];
};
Matrix base_Matrix =  {0,2,0,0,0,0,0,
                       1,1,0,0,0,0,0,
                       0,1,1,0,0,0,0,
                       0,4,4,1,0,0,0,
                       0,6,6,3,1,0,0,
                       0,4,4,3,2,1,0,
                       0,1,1,1,1,1,1};
Matrix mult(Matrix a, Matrix b)
{
    Matrix hh = {0};
    for(int i = 0;i<7;i++)
        for(int j = 0;j<7;j++)
            for(int k = 0;k<7;k++){
                hh.c[i][j] += (a.c[i][k]*b.c[k][j])%mod;
                hh.c[i][j] %= mod;
            }
    return hh;
}
Matrix qpow_Matrix(Matrix a, LL b)
{
    Matrix base = a;
    Matrix ans;
    for(int i = 0;i<7;i++)
        for(int j = 0;j<7;j++)
            if(i==j)    ans.c[i][j] = 1;
            else ans.c[i][j] = 0;

    while(b){
        if(b&1) ans = mult(ans, base);
        base = mult(base, base);
        b>>=1;
    }
    return ans;
}
void init() {

}
void solve() {
    LL n, a, b;
    cin >> n >> a >> b;
    Matrix begin ={a,b,16,8,4,2,1};
    if(n==1){
        cout << a << endl;
        return;
    }
    if(n==2){
        cout << b << endl;
        return;
    }
    Matrix tmp = qpow_Matrix(base_Matrix, n-2);
    Matrix ans = mult(begin, tmp);
//    for(int i = 0;i<7;i++){
//        for(int j = 0;j<7;j++)
//            cout << ans.c[i][j] << " ";
//        cout << endl;
//    }
//    cout << endl;
    cout << ans.c[0][1] << endl;
}
int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
//        freopen("output.txt", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while(T--){
        init();
        solve();
    }
    return 0;
}

　　具体的矩阵构造会出一篇blog详细讲解