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HDU 1757 A Simple Math Problem 矩阵快速幂
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
对于这样一个式子，通过矩阵与线性变换的关系，可以轻松的构造出这样的矩阵
A0:
9
8
7
6
5
4
3
2
1
0
A1:
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
那么f(n)=A1^(n-9)*A0
快速幂即可
#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_backtypedef 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 = 11;LL k, 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) { Matrix ret(a.n, b.m); for(int i = 1; i <= a.n; i++) { for(int j = 1; j <= b.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 pow(Matrix mat, LL p) { if(p == 1) return mat; Matrix ret = pow(mat * mat, p / 2); if(p & 1) ret = ret * mat; return ret;}int main() { int a[10]; while(cin >> k >> mod) { for(int i = 0; i < 10; i++) cin >> a[i]; Matrix A(10, 10), A0(10, 1); if(k < 10) cout << k % mod << endl; else { for(int i = 1; i <= 10; i++) A.data[1][i] = a[i - 1]; for(int i = 2; i <= 10; i++) A.data[i][i - 1] = 1; for(int i = 1, j = 9; i <= 10; i++, j--) A0.data[i][1] = j; A = pow(A, k - 9); A0 = A * A0; cout << A0.data[1][1] << endl; } } return 0;}
HDU 1757 A Simple Math Problem 矩阵快速幂
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