题意: 有$n$种球，每种有无限个，同时第$i$种球有一个代价$c_i$，你要拿不超过$w$个球。如果最后第$i$种球你拿了$k_i$个，那么你会获得$\prod_{1\leq
跑的比较快的多项式板子
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#include <cstring> #include <iostream> #include <random> using int_t = long long int; using std::cin; using std::cout; using std::endl; const int_t LARGE = 2e6; const int_t mod = 998244353; const int_t g = 3; int_t power(int_t b, int_t i) { int_t r = 1; if (i < 0) i = (i % (mod - 1) + mod - 1) % (mod - 1); while (i) { if (i & 1) r = r * b % mod; b = b * b % mod; i >>= 1; } return r; } void makeflip(int_t* arr, int_t size2) { int_t len = (1 << size2); arr[0] = 0; for (int_t i = 1; i < len; i++) { arr[i] = (arr[i >> 1] >> 1) | ((i & 1) << (size2 - 1)); } } int_t upper2n(int_t x) { int_t r = 0; while ((1 << r) < x) r++; return r; } template <int_t arg = 1> void transform(int_t* A, int_t size2, int_t* flip) { int_t len = (1 << size2); for (int_t i = 0; i < len; i++) { int_t r = flip[i]; if (r > i) std::swap(A[i], A[r]); } for (int_t i = 2; i <= len; i *= 2) { int_t mr = power(g, arg * (mod - 1) / i); for (int_t j = 0; j < len; j += i) { int_t curr = 1; for (int_t k = 0; k < i / 2; k++) { int_t u = A[j + k], v = curr * A[j + k + i / 2] % mod; A[j + k] = (u + v) % mod; A[j + k + i / 2] = (u - v + mod) % mod; curr = curr * mr % mod; } } } } void poly_inv(const int_t* A, int_t n, int_t* result) { /* 这里的n是模x^n 计算B(x)*A(x) = 1 mod x^n, 其中A(x)已知 假设已知A(x)*B(x) = 1 mod x^{ceil(n/2)} 假设C(x)*A(x) = 1 mod x^n (A(x)B(x)-1)^2 = A^2(x)B^2(x)-2A(x)B(x)+1= 0 A(x)B^2(x)-2B(x)+C(x) = 0 C(x) = 2B(x)-A(x)B^2(x) */ // #ifdef DEBUG // cout << "at " << n << endl; // #endif if (n == 1) { result[0] = power(A[0], -1); return; } int_t next = n / 2 + n % 2; poly_inv(A, next, result); //次数不要选错了，应该用n次的A和B去卷 int_t size2 = upper2n(n + 2 * next + 1); static int_t X[LARGE]; static int_t Y[LARGE]; int_t len = (1 << size2); // 写错设置范围了！ memset(X + n, 0, sizeof(X[0]) * (len - n)); memset(Y + next, 0, sizeof(Y[0]) * (len - next)); memcpy(X, A, sizeof(A[0]) * n); memcpy(Y, result, sizeof(result[0]) * next); static int_t fliparr[LARGE]; makeflip(fliparr, size2); transform<1>(X, size2, fliparr); transform<1>(Y, size2, fliparr); for (int_t i = 0; i < len; i++) { X[i] = (2 * Y[i] - X[i] * Y[i] % mod * Y[i] % mod + mod) % mod; } transform<-1>(X, size2, fliparr); const int_t inv = power(len, -1); for (int_t i = 0; i < n; i++) result[i] = X[i] * inv % mod; #ifdef DEBUG cout << "poly inv at " << n << endl; cout << "result = "; for (int_t i = 0; i < next; i++) cout << result[i] << " "; cout << endl; #endif } int_t poly_divat(const int_t* F, const int_t* G, int_t n, int_t k) { /* n次多项式F和G 计算F(x)/G(x)的k次项前系数 考虑F(x)*G(-x)/G(x)*G(-x)，分母只有偶数次项，写为C(x^2)；分子写成xA(x^2)+B(x^2)，如果k是奇数，那么递归(A,C,n,k/2)，如果k是偶数，那么递归(B,C,n,k/2) 到k<=n时直接计算 */ int_t size2 = upper2n(2 * n + 1); int_t len = 1 << size2; static int_t fliparr[LARGE]; makeflip(fliparr, size2); static int_t T1[LARGE], T2[LARGE], T3[LARGE]; for (int_t i = 0; i < len; i++) { if (i <= n) T1[i] = F[i], T2[i] = G[i]; else T1[i] = T2[i] = T3[i] = 0; } const int_t inv = power(len, -1); while (k >= n) { #ifdef DEBUG cout << "curr at k = " << k << endl; #endif for (int_t i = 0; i < len; i++) { if (i <= n) { T3[i] = T2[i] * (i % 2 ? (mod - 1) : 1); } else T3[i] = 0; } transform(T1, size2, fliparr); transform(T2, size2, fliparr); transform(T3, size2, fliparr); for (int_t i = 0; i < len; i++) { T1[i] = T1[i] * T3[i] % mod; T2[i] = T2[i] * T3[i] % mod; } transform<-1>(T1, size2, fliparr); transform<-1>(T2, size2, fliparr); #ifdef DEBUG cout << "prod T1 = "; for (int_t i = 0; i < len; i++) cout << T1[i] * inv % mod << " "; cout << endl; cout << "prod T2 = "; for (int_t i = 0; i < len; i++) cout << T2[i] * inv % mod << " "; cout << endl; #endif for (int_t i = 0; i < len; i++) { if (i * 2 < len) { T2[i] = T2[i * 2] * inv % mod; } else T2[i] = 0; } int_t b = k % 2; for (int_t i = 0; i < len; i++) { if (i % 2 == b) { T1[i / 2] = T1[i] * inv % mod; #ifdef DEBUG cout << "at " << i << " assgin " << T1[i] * inv << " to " << i / 2 << endl; #endif } #ifdef DEBUG cout << "assign 0 to " << i << endl; #endif if (i > 0) T1[i] = 0; } #ifdef DEBUG cout << "finished T1 = "; for (int_t i = 0; i < len; i++) cout << T1[i] % mod << " "; cout << endl; cout << "finished T2 = "; for (int_t i = 0; i < len; i++) cout << T2[i] % mod << " "; cout << endl; #endif k >>= 1; } poly_inv(T2, k + 1, T3); // #ifdef DEBUG // cout << "finished k = " << k << endl; // cout << "T1 = "; // for (int_t i = 0; i < len; i++) // cout << T1[i] % mod << " "; // cout << endl; // cout << "T2 = "; // for (int_t i = 0; i < len; i++) // cout << T2[i] % mod << " "; // cout << endl; // cout << "T2 inv = "; // for (int_t i = 0; i < len; i++) // cout << T3[i] % mod << " "; // cout << endl; // #endif int_t result = 0; //计算结果的k次项 for (int_t i = 0; i <= k; i++) { result = (result + T1[i] * T3[k - i] % mod) % mod; } return result; } void poly_mul(const int_t* A, int_t n, const int_t* B, int_t m, int_t* C) { /* 计算n次多项式A与m次多项式B的乘积 */ int_t size2 = upper2n(n + m + 1); int_t len = 1 << size2; static int_t T1[LARGE], T2[LARGE]; for (int_t i = 0; i < len; i++) { if (i <= n) T1[i] = A[i]; else T1[i] = 0; if (i <= m) T2[i] = B[i]; else T2[i] = 0; } static int_t fliparr[LARGE]; makeflip(fliparr, size2); transform(T1, size2, fliparr); transform(T2, size2, fliparr); for (int_t i = 0; i < len; i++) T1[i] = T1[i] * T2[i] % mod; transform<-1>(T1, size2, fliparr); int_t inv = power(len, -1); for (int_t i = 0; i <= n + m; i++) C[i] = T1[i] * inv % mod; } int_t poly_linear_rec(const int_t* A0, const int_t* F0, int_t n, int_t k) { /* 计算线性递推 F[1],F[2]...F[k] 递推系数 A[0],A[1]...A[k-1] 首项 A[m]=A[m-1]*F[1]+A[m-2]*F[2]+...+A[m-k]*F[k] */ static int_t T1[LARGE], T2[LARGE]; static int_t Ax[LARGE], Fx[LARGE]; Fx[0] = 0; Ax[k] = 0; for (int i = 1; i <= k; i++) { Fx[i] = F0[i]; Ax[i - 1] = A0[i - 1]; } poly_mul(Ax, k, Fx, k, T1); T1[0] = Ax[0]; for (int i = 1; i <= k - 1; i++) { T1[i] = (Ax[i] - T1[i] + mod) % mod; } for (int i = k; i <= 2 * k; i++) T1[i] = 0; T2[0] = 1; for (int i = 1; i <= k; i++) T2[i] = (mod - Fx[i]) % mod; // #ifdef DEBUG // cout << "T1 = "; // for (int_t i = 0; i <= k; i++) { // cout << T1[i] << " "; // } // cout << endl; // cout << "T2 = "; // for (int_t i = 0; i <= k; i++) { // cout << T2[i] << " "; // } // cout << endl; // #endif return poly_divat(T1, T2, k, n); } void poly_log(const int_t* A, int_t n, int_t* out) { /* 计算log(A(x)), A(x)为n次多项式 DlogF(x) =DF(x)/F(x) */ static int_t Ad[LARGE]; static int_t Finv[LARGE]; static int_t R[LARGE]; const int_t m = n - 1; for (int_t i = 0; i <= m; i++) { Ad[i] = A[i + 1] * (i + 1) % mod; } Ad[n] = 0; poly_inv(A, n + 1, Finv); poly_mul(Ad, m, Finv, n, R); for (int_t i = 1; i <= n; i++) { out[i] = R[i - 1] * power(i, -1) % mod; } } void poly_exp(const int_t* A, int_t n, int_t* out) { /* 计算exp(A(x)), A(x)为n次多项式 H(x)=G(x)(1-logG(x)+A(x)) H(x)为当次递推项，G(x)为上一次递推项 */ if (n == 1) { out[0] = 1; return; } int_t r = n / 2 + n % 2; poly_exp(A, r, out); static int_t G[LARGE]; static int_t G2[LARGE]; static int_t logG[LARGE]; static int_t R[LARGE]; for (int_t i = 0; i < r; i++) G[i] = out[i]; for (int_t i = r; i < n; i++) G[i] = 0; poly_log(G, n - 1, logG); // for (int_t i = r; i < n; i++) // logG[i] = 0; for (int_t i = 0; i < n; i++) { G2[i] = (mod - logG[i] + A[i]) % mod; } G2[0] = (G2[0] + 1) % mod; poly_mul(G, n - 1, G2, n - 1, R); for (int_t i = 0; i < n; i++) out[i] = R[i]; #ifdef DEBUG cout << "at " << n << endl; cout << "A = "; for (int_t i = 0; i < n; i++) cout << A[i] << " "; cout << endl; cout << "G = "; for (int_t i = 0; i < n; i++) cout << G[i] << " "; cout << endl; cout << "logG = "; for (int_t i = 0; i < n; i++) cout << logG[i] << " "; cout << endl; cout << "G2 = "; for (int_t i = 0; i < n; i++) cout << G2[i] << " "; cout << endl; cout << "R = "; for (int_t i = 0; i < n; i++) cout << R[i] << " "; cout << endl; cout << endl; #endif } int main() { std::ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); int_t n; cin >> n; static int_t A[LARGE], B[LARGE]; static int_t C[LARGE]; for (int_t i = 0; i < n; i++) cin >> A[i]; poly_exp(A, n, B); for (int_t i = 0; i < n; i++) cout << B[i] << " "; return 0; } |
常系数线性递推的新做法 – 计算[x^k]P(x)/Q(x)
$$ a_n=\sum_{1\le i\le k}{f_ia_{n-i}} \\ \left\{ a_0,a_1....a_{k-1} \right\} \text{已知} \\ \\ \text{设}F\left(
Nginx路由匹配模拟器
https://nginx.viraptor.info/
真是我的救星，调路由调吐了
CFGYM103104I-WHUPC I Sequence
CF1521C Nastia and a Hidden Permutation
真是个沙比提，比赛的时候也想出来了，但是因为太迷糊了没写完
CFGYM102394L CCPC2019哈尔滨 L题 LRU Algorithm
CFGYM102394E CCPC2019哈尔滨 E题 Exchanging Gifts
假设我们能维护出最终序列的长度L和最终序列出现最多的数的个数cnt，假设$cnt\leq\frac L 2$ ，那么答案是L,这时候我们把序列升序和降序对起来就构造出n的结果。
假设$cnt>\frac L 2$,那么答案是$2(L-cnt)$，我们仍然把序列升序和降序对应起来，答案是$L-(cnt-(L-cnt))=2*(L-cnt)$
比如
1 2 3 3 3 3 3
3 3 3 3 3 2 1
中间有3个3是重了的，那么重的部分有cnt-(L-cnt)个，其中L-cnt表示的是1 2的长度，所以总答案是L-(cnt-(L-cnt))
现在的问题在于如何维护出这个众数，并且判定$cnt\leq \frac L 2$是否成立。
考虑一种线性时间求求序列众数(出现次数大于一半)的方法：
令f(i)表示我们从头开始扫到第i个元素的时候(用$a_i$表示)，序列中出现次数最多的数与其他的数出现次数之和的差值。同时我们需要维护这个数是多少(用x来表示)。
每次扫到$a_i$的时候:
- 如果$a_i=x$，那么f(i)=f(i-1)+1,x不变
- 如果$a_i\neq x$,那么f(i)=f(i-1)-1，然后如果$f(i)<0$,那么x变为$a_i$，同时$f(i)$取反。
对于这个题，如果我们要使用这种求众数的方法，核心在于如何考虑操作2（合并两个序列时）如何处理。
我们维护每个序列的f和x，合并两个序列的时候：
- 如果他们的f相同，那么新序列的f不变，x相加。
- 如果他们的f不同，那么考虑下两个序列的哪一个的x比较大。如果他们相同，那么新序列的f就写为0(这时候可能存在两个众数)，x从原来的x里随便选一个。如果他们不同，那x选择为f较大的那一个，同时f设置为较大值减掉较小值。
所以对于这个题，我们先用这种求众数的方法算出来最终序列的众数。
但此时求出来的众数，仅仅是在保证存在一个出现次数超过序列长度一半时候的众数，如果不存在这样子的众数，也会得出来一个结果，但是并不具有意义。
所以我们再跑一遍递推，求出来我们上一步求的众数的出现次数，然后我们就可以照着前文所述来算答案了。
*此外，这个题卡常*
我跑了半天性能分析后大概知道了几个卡常的点:
1. 输入量非常巨大，请考虑一次性读入输入数据后自行用缓冲区处理输入。
2. std::vector的构造函数非常慢(性能分析显示,$10^6$次调用大约花了80ms)，所以不要使用vector来存储变长的序列，考虑自行分配-回收内存。
3. State结构体的构造函数占了大概40ms的时间，考虑强制内联。
另外，读入函数千万不要写错了！不要写成chr>='9'！
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#pragma GCC optimize("O2") #include <assert.h> #include <inttypes.h> #include <algorithm> #include <iostream> #include <vector> using int_t = int; using std::cin; using std::cout; using std::endl; const int_t LARGE = 1e6; char inputbuf[(int(2e8) / 1024) * 1024]; char* head = inputbuf; inline char nextchar() { return *(head++); } void initinput() { fread(inputbuf, 1024, sizeof(inputbuf) / 1024, stdin); } struct State { int64_t len; int_t mostval; int64_t mostcount; inline State(int64_t len = 0, int_t mostval = 0, int64_t mostcount = 0) : len(len), mostval(mostval), mostcount(mostcount) {} State operator+(const State& rhs) const { State result(len + rhs.len, 0, 0); if (mostval == rhs.mostval) result.mostval = mostval, result.mostcount = mostcount + rhs.mostcount; else { if (mostcount > rhs.mostcount) { result.mostcount = mostcount - rhs.mostcount; result.mostval = mostval; } else { result.mostcount = rhs.mostcount - mostcount; result.mostval = rhs.mostval; } } return result; } }; std::ostream& operator<<(std::ostream& os, const State& state) { os << "State{len=" << state.len << ",mostval=" << state.mostval << ",mostcount=" << state.mostcount << "}"; return os; } struct Opt { int_t type; int_t* data; int_t datalen; int_t x1, x2; int64_t mostcount = 0; } opts[LARGE + 1]; State dp[LARGE + 1]; int_t n; template <class T> void read(T& x) { x = 0; char chr = nextchar(); while (chr < '0' || chr > '9') chr = nextchar(); while (chr >= '0' && chr <= '9') { x = x * 10 + chr - '0'; chr = nextchar(); } assert(x >= 0); } template <class T> void write(T x) { assert(x >= 0); if (x > 9) write(x / 10); putchar('0' + x % 10); } int main() { // freopen("input.txt", "r", stdin); initinput(); int_t T; read(T); while (T--) { read(n); for (int_t i = 1; i <= n; i++) { auto& ref = opts[i]; if (ref.data) { delete[] ref.data; ref.data = nullptr; } dp[i] = State(); read(ref.type); if (ref.type == 1) { int_t k; read(k); int_t sum = 0, val = 0; ref.data = new int_t[k + 1]; ref.datalen = k; for (int_t i = 1; i <= k; i++) { int_t x; read(x); ref.data[i] = x; if (x == val) sum++; else sum--; if (sum < 0) { sum *= -1, val = x; } } // ref.seq.shrink_to_fit(); dp[i] = State(k, val, sum); } else { read(ref.x1); read(ref.x2); } } for (int_t i = 1; i <= n; i++) { const auto& ref1 = opts[i]; if (ref1.type == 2) { dp[i] = dp[ref1.x1] + dp[ref1.x2]; } } #ifdef DEBUG for (int_t i = 1; i <= n; i++) { cout << "dp " << i << " = " << dp[i] << endl; } #endif int_t mostval = dp[n].mostval; for (int_t i = 1; i <= n; i++) { auto& ref = opts[i]; if (ref.type == 1) ref.mostcount = std::count(ref.data + 1, ref.data + 1 + ref.datalen, mostval); else ref.mostcount = opts[ref.x1].mostcount + opts[ref.x2].mostcount; } int64_t mostcount = opts[n].mostcount; #ifdef DEBUG cout << "final len " << dp[n].len << endl; cout << "mostcount " << mostcount << endl; cout << "mostval " << mostval << endl; #endif if (mostcount * 2 <= dp[n].len) { write(dp[n].len); } else { write(2 * (dp[n].len - mostcount)); } puts(""); } return 0; } |
CFGYM102394I CCPC2019哈尔滨 I题 Interesting Permutations
签到题我都不会做，我爬了。
首先检测一些明显非法的情况:
CFGYM102394A CCPC2019哈尔滨A Artful Paintings
一眼看过去是差分约束板子题，实际上也就是差分约束板子题。