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贴一个一般般的代码#include <iostream> #include <algorithm> using namespace std; int mA[32][3][3];//幂矩阵 int tA[3][3];//临时存储矩阵 int A[3];//A[0] 均为偶数的方案数 A[1] 一奇一偶的方案数 A[2] 均为奇数的方案数 int main() { int T, N, sum; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { mA[0][i][j] = 1; } } mA[1][0][0] = 2; mA[1][0][1] = 1; mA[1][0][2] = 0; mA[1][1][0] = 2; mA[1][1][1] = 2; mA[1][1][2] = 2; mA[1][2][0] = 0; mA[1][2][1] = 1; mA[1][2][2] = 2; for (int i = 2; i < 32; i++) { for (int k1 = 0; k1 < 3; k1++) { for (int k2 = 0; k2 < 3; k2++) { sum = 0; for (int j = 0; j < 3; j++) { sum += mA[i - 1][k1][j] * mA[i - 1][j][k2]; } tA[k1][k2] = sum % 10007; } } for (int k1 = 0; k1 < 3; k1++) { for (int k2 = 0; k2 < 3; k2++) { mA[i][k1][k2] = tA[k1][k2]; } } } cin >> T; for (int so = 0; so < T; so++) { int i = 0; A[0] = 1; A[1] = 0; A[2] = 0; cin >> N; while (N) { i++; if (N & true) { for (int k1 = 0; k1 < 3; k1++) { sum = 0; for (int j = 0; j < 3; j++) { sum += A[j] * mA[i][k1][j]; } tA[0][k1] = sum % 10007; } for (int k1 = 0; k1 < 3; k1++) { A[k1] = tA[0][k1]; } } N >>= 1; } cout << A[0] << '\n'; } return 0; } Followed by: Post your reply here: |
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