HDU 5950 Recursive sequence (矩阵快速幂)

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#矩阵快速幂

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本文探讨了递归数列的高效求解方法，通过矩阵快速幂技术解决了一个特定的递归数列问题，该数列包含了一次项、平方项、三次方项和四次方项的复杂组合。

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4283 Accepted Submission(s): 1852

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 i^4. 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 < 2^31 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）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5950

题目大意：给f[1]，f[2]和递推式f[i]=2*f[i-2]+f[i-1]+i^4，求f[n]

题目分析：看到数据范围和递推式第一反应就是矩阵快速幂，f数组好办，i^4转移需先将(i+1)^4展开

(i+1)^4=i^4+4i^3+6i^2+4i+1，需要维护一个7*7的转移矩阵transation

（1，1，0，0，0，0，0）

（2，0，0，0，0，0，0）

（1，0，1，0，0，0，0）

（4，0，4，1，0，0，0）

（6，0，6，3，1，0，0）

（4，0，4，3，2，1，0）

（1，0，1，1，1，1，1）

（f[i]，f[i-1]，i^4，i^3，i^2，i，1）* transation =（f[i+1]，f[i]，(i+1)^4，(i+1)^3，(i+1)^2，i+1，1）

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const N = 7;
ll const MOD = 2147493647;
 
ll n, a, b;
 
struct matrix {
    ll mat[N][N];
    matrix() {
        memset(mat, 0, sizeof(mat));
    }
};
    
matrix mmul(matrix a, matrix b) {
    matrix ans;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (a.mat[i][j]) {
                for (int k = 0; k < N; k++) {
                    ans.mat[i][k] += (a.mat[i][j] * b.mat[j][k]) % MOD;
                    ans.mat[i][k] %= MOD;
                }
            }
        }
    }
    return ans;
}
 
matrix mqpow(matrix a, ll n) {
    matrix ans;
    for (int i = 0; i < N; i++) {
        ans.mat[i][i] = 1;
    }
    while (n) {
        if (n & 1) {
            ans = mmul(ans, a);
        }
        a = mmul(a, a);
        n >>= 1;
    }
    return ans;
}
 
int main() {
    int T;
    scanf("%d", &T);
    matrix x;
    while (T--) {
        ll n, a, b;
        scanf("%lld %lld %lld", &n, &a, &b);
        if (n == 1) {
            printf("%lld\n", a % MOD);
            continue;
        }
        if (n == 2) {
            printf("%lld\n", b % MOD);
            continue;
        }
        memset(x.mat, 0, sizeof(x.mat));
        x.mat[0][0] = 1; x.mat[0][1] = 1;
        x.mat[1][0] = 2;
        x.mat[2][0] = 1; x.mat[2][2] = 1;
        x.mat[3][0] = 4; x.mat[3][2] = 4; x.mat[3][3] = 1;
        x.mat[4][0] = 6; x.mat[4][2] = 6; x.mat[4][3] = 3; x.mat[4][4] = 1;
        x.mat[5][0] = 4; x.mat[5][2] = 4; x.mat[5][3] = 3; x.mat[5][4] = 2; x.mat[5][5] = 1;
        x.mat[6][0] = 1; x.mat[6][2] = 1; x.mat[6][3] = 1; x.mat[6][4] = 1; x.mat[6][5] = 1; x.mat[6][6] = 1;
        x = mqpow(x, n - 2);
        ll init[7] = {b, a, 16, 8 ,4, 2, 1};
        ll ans = 0;
        for (int i = 0; i < N; i++) {
            ans = ans + init[i] * x.mat[i][0];
            ans %= MOD;
        }
        printf("%lld\n", ans);
    }
}