链接:
http://acm.hdu.edu.cn/showproblem.php?pid=6030
题意:
给一个手链染色,每连续素数个数的珠子中红色不能比蓝的多,问有多少种情况
题解:
公式为f[i]=f[i-1]+f[i-3],类似菲波那切数列,使用矩阵快速幂即可
代码:
31 typedef vector<ll> vec; 32 typedef vector<vec> mat; 33 34 mat mul(mat &A, mat &B) { 35 mat C(A.size(), vec(B[0].size())); 36 rep(i, 0, A.size()) rep(k, 0, B.size()) rep(j, 0, B[0].size()) 37 C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD; 38 return C; 39 } 40 41 mat pow(mat A, ll n) { 42 mat B(A.size(), vec(A.size())); 43 rep(i, 0, A.size()) B[i][i] = 1; 44 while (n>0) { 45 if (n & 1) B = mul(B, A); 46 A = mul(A, A); 47 n >>= 1; 48 } 49 return B; 50 } 51 52 int main() { 53 int T; 54 cin >> T; 55 while (T--) { 56 mat A(3, vec(3)); 57 A[0][0] = 1, A[0][1] = 0, A[0][2] = 1; 58 A[1][0] = 1, A[1][1] = 0, A[2][2] = 0; 59 A[2][0] = 0, A[2][1] = 1, A[2][2] = 0; 60 ll n; 61 cin >> n; 62 A = pow(A, n - 2); 63 ll ans = (A[2][0] * 6 + A[2][1] * 4 + A[2][2] * 3) % MOD; 64 cout << ans << endl; 65 } 66 return 0; 67 }
时间: 2024-10-19 10:39:21