首先fib数列可以很随意的推出来矩阵解法,然后这里就是要处理一个关于矩阵的等比数列求和的问题,这里有一个logn的解法,类似与这样
A^0+A^1+A^2+A^3 = A^0 + A^1 + A^2 * (A^0 + A^1) 处理就好了。
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack> #include <map> #include <set> #include <climits> #include <iostream> #include <string> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<double, double> PDD; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 6; LL k, b, n, mod; struct Matrix { int n, m; LL data[maxn][maxn]; Matrix(int n = 0, int m = 0): n(n), m(m) { memset(data, 0, sizeof(data)); } }; Matrix operator * (Matrix a, Matrix b) { int n = a.n, m = b.m; Matrix ret(n, m); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { for(int k = 1; k <= a.m; k++) { ret.data[i][j] += a.data[i][k] * b.data[k][j]; ret.data[i][j] %= mod; } } } return ret; } Matrix operator + (Matrix a, Matrix b) { for(int i = 1; i <= a.n; i++) { for(int j = 1; j <= a.m; j++) { a.data[i][j] += b.data[i][j]; a.data[i][j] %= mod; } } return a; } Matrix pow(Matrix mat, LL k) { if(k == 0) { Matrix ret(mat.n, mat.m); for(int i = 1; i <= mat.n; i++) ret.data[i][i] = i; return ret; } if(k == 1) return mat; Matrix ret = pow(mat * mat, k / 2); if(k & 1) ret = ret * mat; return ret; } Matrix calc(Matrix mat, LL p) { if(p == 0) return pow(mat, b); if(p == 1) return pow(mat, b) + pow(mat, k + b); int midval = (p + 1) % 2 == 0 ? (p / 2) : (p / 2) - 1; Matrix ret = calc(mat, midval); ret = ret + pow(mat, (midval + 1) * k) * ret; if((p + 1) & 1) ret = ret + pow(mat, p * k + b); return ret; } int main() { while(cin >> k >> b >> n >> mod) { Matrix A(2, 2); A.data[1][1] = A.data[1][2] = A.data[2][1] = 1; A.data[2][2] = 0; Matrix ans = calc(A, n - 1); cout << ans.data[2][1] << endl; } return 0; }
时间: 2024-11-08 20:26:21