首先把递推式求解转化为矩阵求幂，再利用特征多项式f(λ)满足f(A) = 0,将矩阵求幂转化为多项式相乘，
最后利用傅里叶变换的高效算法（迭代取代递归）（参见算法导论）解决。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <cmath> 9 #include <ctime> 10 #pragma comment(linker, "/STACK:102400000,102400000") 11 using namespace std; 12 #define lson (u << 1) 13 #define rson (u << 1 | 1) 14 typedef long long ll; 15 const double eps = 1e-6; 16 const double pi = acos(-1.0); 17 const int maxn = 4e4 + 10; 18 const int maxm = 1050; 19 const int mod = 119; 20 const int inf = 0x3f3f3f3f; 21 22 int n, a, b, p, q; 23 int size; 24 int f[maxn], g[maxn]; 25 26 struct Complex{ 27 double ii, ij;//ii::real, ij::image 28 Complex(double ii = 0, double ij = 0) : ii(ii), ij(ij) {} 29 // Complex clear() { this->ii = this->ij = 0; } 30 Complex operator + (const Complex &rhs) const{ 31 return Complex(ii + rhs.ii, ij + rhs.ij); 32 } 33 Complex operator - (const Complex &rhs) const{ 34 return Complex(ii - rhs.ii, ij - rhs.ij); 35 } 36 Complex operator * (const Complex &rhs) const{ 37 return Complex(ii * rhs.ii - ij * rhs.ij, ii * rhs.ij + ij * rhs.ii); 38 } 39 }; 40 41 Complex a1[maxn], a2[maxn]; 42 43 void fft(Complex *src, int len, int rev){ 44 //len is power of 2 45 //rev == 1::dft rev == -1::idft 46 for(int i = 1, j = 0; i < len; i++){ 47 for(int k = len >> 1; k > (j ^= k); k >>= 1) ; 48 if(i < j) swap(src[i], src[j]); 49 } 50 for(int i = 2; i <= len; i <<= 1){ 51 Complex wi(cos(2 * pi * rev / i), sin(2 * pi * rev / i)); 52 //(wi)^i = 1 53 for(int j = 0; j < len; j += i){ 54 //using iteration insetad of recursion 55 Complex w(1.0, 0.0); 56 //w = (wi)^0 57 for(int k = j; k < j + i / 2; k++){ 58 Complex tem = w * src[k + i / 2]; 59 src[k + i / 2] = src[k] - tem; 60 src[k] = src[k] + tem; 61 w = w * wi; 62 } 63 } 64 } 65 if(rev == -1){ 66 for(int i = 0; i < len; i++) src[i].ii = (src[i].ii / len + eps); 67 } 68 } 69 70 void multi(int *src1, int *src2, int len){ 71 for(int i = 0; i < len; i++){ 72 a1[i].ii = a1[i].ij = a2[i].ii = a2[i].ij = 0; 73 if(i < q){ 74 a1[i].ii = (double)src1[i]; 75 a2[i].ii = (double)src2[i]; 76 } 77 } 78 fft(a1, len, 1), fft(a2, len, 1); 79 for(int i = 0; i < len; i++) a1[i] = a1[i] * a2[i]; 80 fft(a1, len, -1); 81 for(int i = 0; i < len; i++) g[i] = (int)((ll)(a1[i].ii + eps) % mod); 82 for(int i = 2 * q - 2; i >= q; i--){ 83 //this is because for the fisrt row in matrix A, 84 //which satisfies ths (f(n + q),...,f(n))T = A((f(n + q - 1),...,f(n - 1))T) 85 //only two elements are nonzero integers 86 g[i - q] = (g[i - q] + g[i] * b) % mod; 87 g[i - p] = (g[i - p] + g[i] * a) % mod; 88 } 89 memcpy(src1, g, sizeof(int) * q); 90 } 91 92 int tmp[maxn], ans[maxn]; 93 94 int main(){ 95 //freopen("in.txt", "r", stdin); 96 while(~scanf("%d%d%d%d%d", &n, &a, &b, &p, &q)){ 97 a %= mod, b %= mod; 98 f[0] = 1; 99 for(int i = 1; i < q; i++){ 100 f[i] = i < p ? a + b : a * f[i - p] + b; 101 f[i] %= mod; 102 } 103 if(n < q){ 104 printf("%d ", f[n]); 105 continue; 106 } 107 size = 1; 108 while(size <= (q - 1) * 2) size <<= 1; 109 memset(tmp, 0, sizeof tmp); 110 memset(ans, 0, sizeof ans); 111 ans[0] = tmp[1] = 1; 112 while(n){ 113 if(n & 1) multi(ans, tmp, size); 114 multi(tmp, tmp, size); 115 n >>= 1; 116 } 117 int res = 0; 118 for(int i = 0; i < q; i++){ 119 res = (res + ans[i] * f[i]) % mod; 120 } 121 printf("%d ", res); 122 } 123 return 0; 124 }