UVa 10870 (矩阵快速幂) Recurrences
2015-03-14 13:53
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给出一个d阶线性递推关系,求f(n) mod m的值。
代码君
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn = 20; typedef long long Matrix[maxn][maxn]; typedef long long Vector[maxn]; int d, n, m; void matrix_mul(Matrix A, Matrix B, Matrix res) { Matrix C; memset(C, 0, sizeof(C)); for(int i = 0; i < d; i++) for(int j = 0; j < d; j++) for(int k = 0; k < d; k++) C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % m; memcpy(res, C, sizeof(C)); } void matrix_pow(Matrix A, int n, Matrix res) { Matrix a, r; memcpy(a, A, sizeof(a)); memset(r, 0, sizeof(r)); for(int i = 0; i < d; i++) r[i][i] = 1; while(n) { if(n&1) matrix_mul(r, a, r); n >>= 1; matrix_mul(a, a, a); } memcpy(res, r, sizeof(r)); } int main() { //freopen("in.txt", "r", stdin); while(cin >> d >> n >> m && d) { Matrix A; memset(A, 0, sizeof(A)); Vector a, f; for(int i = 0; i < d; i++) { cin >> a[i]; a[i] %= m; } for(int i = 0; i < d; i++) { cin >> f[i]; f[i] %= m; } if(n <= d) { cout << f[n-1] << "\n"; continue; } for(int i = 0; i < d-1; i++) A[i][i+1] = 1; for(int i = 0; i < d; i++) A[d-1][i] = a[d-i-1]; matrix_pow(A, n-d, A); long long ans = 0; for(int i = 0; i < d; i++) ans = (ans + A[d-1][i]*f[i]) % m; cout << ans << "\n"; } return 0; }
代码君
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