Product Oriented Recurrence

先化簡原式子

c ^ x * f[x]? = c ^ (x-1) * f[x-1] * c ^ (x-2) * f[x-2] * c ^ (x-3) * f[x-3]

及g[x] = c ^ x * f[x]

g[x] = g[x-1] * g[x-2] * g[x-3]

然后用矩陣快速冪計算g1, g2, g3的貢獻， 計算出gn 之后 轉回 fn

//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0);using namespace std;const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1);template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());LL power(LL a, LL b) {LL ans = 1;while(b) {if(b & 1) ans = ans * a % mod;a = a * a % mod; b >>= 1;}return ans; }int MOD = (int)1e9 + 6;struct Matrix {int a[3][3];Matrix() {memset(a, 0, sizeof(a));}void init() {for(int i = 0; i < 3; i++) {a[i][i] = 1;}}Matrix operator * (const Matrix &B) const {Matrix C;for(int i = 0; i < 3; i++) {for(int j = 0; j < 3; j++) {for(int k = 0; k < 3; k++) {C.a[i][j] += 1LL * a[i][k] * B.a[k][j] % MOD;if(C.a[i][j] >= MOD) C.a[i][j] -= MOD;}}}return C;}Matrix operator ^ (LL b) {Matrix C; C.init();Matrix A = (*this);while(b) {if(b & 1) C = C * A;A = A * A; b >>= 1;}return C;} } M;int mat[3][3] {{1, 1, 1},{1, 0, 0},{0, 1, 0} };LL n, f1, f2, f3, c;int main() {for(int i = 0; i < 3; i++) {for(int j = 0; j < 3; j++) {M.a[i][j] = mat[i][j];}}scanf("%lld%lld%lld%lld%lld", &n, &f1, &f2, &f3, &c);Matrix ret = M ^ (n - 3);f1 = f1 * power(c, 1) % mod;f2 = f2 * power(c, 2) % mod;f3 = f3 * power(c, 3) % mod;LL ans = 1;LL cnt1 = ret.a[0][2];LL cnt2 = ret.a[0][1];LL cnt3 = ret.a[0][0];ans = ans * power(f1, cnt1) % mod;ans = ans * power(f2, cnt2) % mod;ans = ans * power(f3, cnt3) % mod;LL inv = power(power(c, n), mod - 2);ans = ans * inv % mod;printf("%lld\n", ans);return 0; }/* */

?

轉載于:https://www.cnblogs.com/CJLHY/p/11223481.html

總結

以上是生活随笔為你收集整理的Codeforces 1182E Product Oriented Recurrence 矩阵快速幂的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。