5950 Recursive sequence (矩阵快速幂)

题意：递推公式 Fn = Fn-1 + 2 * Fn-2 + n*n，让求 Fn;

析：很明显的矩阵快速幂，因为这个很像Fibonacci数列，所以我们考虑是矩阵，然后我们进行推公式，因为这样我们是无法进行运算的。好像有的思路，最后也没想出来，还是参考的大牛的博客

http://blog.csdn.net/spring371327/article/details/52973534

那是讲的很详细了，就不多说了，注意这个取模不是1e9+7，一开始忘了。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 2147493647;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Matrix{
    LL a[7][7];
    Matrix operator * (const Matrix &p){
        Matrix res;
        for(int i = 0; i < 7; ++i)
            for(int j = 0; j < 7; ++j){
                res.a[i][j] = 0;
                for(int k = 0; k < 7; ++k)
                    res.a[i][j] = (res.a[i][j] + a[i][k] * p.a[k][j]) % mod;
            }
        return res;
    }
};

Matrix quick_pow(Matrix b, LL n){
    Matrix res;
    memset(res.a, 0, sizeof res.a);
    for(int i = 0; i < 7; ++i)  res.a[i][i] = 1;
    while(n){
        if(n & 1)  res = res * b;
        b = b * b;
        n >>= 1;
    }
    return res;
}

int main(){
    Matrix x;
    memset(x.a, 0, sizeof x.a);
    x.a[0][0] = 1;  x.a[0][1] = 2; x.a[0][2] = 1; x.a[0][3] = 4;  x.a[0][4] = 6;
    x.a[0][5] = 4;  x.a[0][6] = 1; x.a[1][0] = 1; x.a[2][2] = 1;  x.a[2][3] = 4;
    x.a[2][4] = 6;  x.a[2][5] = 4; x.a[2][6] = 1; x.a[3][3] = 1;  x.a[3][4] = 3;
    x.a[3][5] = 3;  x.a[3][6] = 1; x.a[4][4] = 1; x.a[4][5] = 2;  x.a[4][6] = 1;
    x.a[5][5] = 1;  x.a[5][6] = 1; x.a[6][6] = 1;
    int T;  cin >> T;
    while(T--){
        LL n, a, b;
        scanf("%I64d %I64d %I64d", &n, &a, &b);
        if(1 == n)  printf("%I64d\n", a);
        else if(2 == n)  printf("%I64d\n", b);
        else{
            Matrix res = quick_pow(x, n-2);
            LL ans = 0;
            ans = (ans + res.a[0][0] * b) % mod;
            ans = (ans + res.a[0][1] * a) % mod;
            ans = (ans + res.a[0][2] * 16) % mod;
            ans = (ans + res.a[0][3] * 8) % mod;
            ans = (ans + res.a[0][4] * 4) % mod;
            ans = (ans + res.a[0][5] * 2) % mod;
            ans = (ans + res.a[0][6]) % mod;
            printf("%I64d\n", ans);
        }
    }
    return 0;
}