网站想上线怎么做,网上做调查问卷赚钱的网站,微信账号注册官网,如何加入wordpressP4705 玩游戏 有ansk∑i1n∑j1m(aibj)knm先舍弃nm不管ansk∑r0k∑i1n∑j1mCkrairbjk−r∑r0k∑i1n∑j1mk!r!(k−r)!airbjk−rk!∑r0k(1r!∑i1nair)(1(k−r)!∑j1mbjk−r)不难发现这就是一个卷积的形式了#xff0c;但是我们现在还不知道∑i1nair,∑i1mbir,(r∈[0,k])设A(x)为∑…P4705 玩游戏 有ansk∑i1n∑j1m(aibj)knm先舍弃nm不管ansk∑r0k∑i1n∑j1mCkrairbjk−r∑r0k∑i1n∑j1mk!r!(k−r)!airbjk−rk!∑r0k(1r!∑i1nair)(1(k−r)!∑j1mbjk−r)不难发现这就是一个卷积的形式了但是我们现在还不知道∑i1nair,∑i1mbir,(r∈[0,k])设A(x)为∑i1nair的生成函数同理可得B(x)A(x)∑r≥0xr∑i1nair∑i1n∑r≥0xrair∑i1n11−aixn−x∑i1n−ai1−aixn−x∑i1n(ln⁡(1−aix))′n−x(ln⁡∏i1n(1−aix))′有ans_k \frac{\sum\limits_{i 1} ^{n} \sum\limits_{j 1} ^{m} (a_i b_j) ^ k}{nm}\\ 先舍弃nm不管\\ ans_k \sum_{r 0} ^{k} \sum_{i 1} ^{n} \sum_{j 1} ^{m} C_{k} ^{r} a_i ^ rb_j ^{k - r}\\ \sum_{r 0} ^{k} \sum_{i 1} ^{n} \sum_{j 1} ^{m} \frac{k!}{r!(k - r)!}a_i ^{r}b_{j} ^{k - r}\\ k!\sum_{r 0} ^{k} \left(\frac{1}{r!} \sum_{i 1} ^{n}a_i ^ r\right) \left(\frac{1}{(k - r)!} \sum_{j 1} ^{m} b_j ^ {k - r}\right)\\ 不难发现这就是一个卷积的形式了但是我们现在还不知道\sum_{i 1} ^{n} a_i ^ r, \sum_{i 1} ^{m} b_i ^ r, (r \in[0, k])\\ 设A(x)为\sum_{i 1} ^{n} a_i ^ r的生成函数同理可得B(x)\\ A(x) \sum_{r \geq 0} x ^ r \sum_{i 1} ^{n} a_i ^ r\\ \sum_{i 1} ^{n} \sum_{r \geq 0} x ^ r a_i ^ r\\ \sum_{i 1} ^{n} \frac{1}{1 - a_i x}\\ n - x \sum_{i 1} ^{n} \frac{-a_i}{1 - a_i x}\\ n - x \sum_{i 1} ^{n} (\ln(1 - a_ix))\\ n - x(\ln \prod_{i 1} ^{n}(1 - a_ix))\\ 有ansk​nmi1∑n​j1∑m​(ai​bj​)k​先舍弃nm不管ansk​r0∑k​i1∑n​j1∑m​Ckr​air​bjk−r​r0∑k​i1∑n​j1∑m​r!(k−r)!k!​air​bjk−r​k!r0∑k​(r!1​i1∑n​air​)((k−r)!1​j1∑m​bjk−r​)不难发现这就是一个卷积的形式了但是我们现在还不知道i1∑n​air​,i1∑m​bir​,(r∈[0,k])设A(x)为i1∑n​air​的生成函数同理可得B(x)A(x)r≥0∑​xri1∑n​air​i1∑n​r≥0∑​xrair​i1∑n​1−ai​x1​n−xi1∑n​1−ai​x−ai​​n−xi1∑n​(ln(1−ai​x))′n−x(lni1∏n​(1−ai​x))′ 对于∏i1n(1−aix)∏i1mid(1−aix)∏imid1n(1−aix)\prod\limits_{i 1} ^{n} (1 - a_ix) \prod\limits_{i 1} ^{mid}(1 - a_ix) \prod\limits_{i mid 1} ^{n} (1 - a_ix)i1∏n​(1−ai​x)i1∏mid​(1−ai​x)imid1∏n​(1−ai​x)所以递归合并复杂度O(nlog⁡nlog⁡n)O(n \log n \log n)O(nlognlogn) 然后再对上面求得的取对数ln⁡\lnln复杂度O(nlog⁡n)O(n \log n)O(nlogn)求一次导可得A(x)A(x)A(x)系数然后再乘上inv[i!]inv[i!]inv[i!]再做一次NTTNTTNTT即可得到答案。 这道题还真是有亿点点细节呀…… #include bits/stdc.husing namespace std;const int mod 998244353, inv2 mod 1 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r 0, int _i 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex a, Complex b) {return Complex((1ll * a.r * b.r % mod 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans Complex(1, 0);while (n) {if (n 1) {ans ans * a;}a a * a;n 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n 0) {return 0;}if(quick_pow(n, (mod - 1) 1).r mod - 1) {return -1;}uniform_int_distributionint r(0, mod - 1);int a r(e);while(quick_pow((1ll * a * a % mod - n mod) % mod, (mod - 1) 1).r 1) {a r(e);}I2 (1ll * a * a % mod - n mod) % mod;int x quick_pow(Complex(a, 1), (mod 1) 1).r, y mod - x;if(x y) swap(x, y);return x;} }const int N 1e6 10;int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];int quick_pow(int a, int n) {int ans 1;while (n) {if (n 1) {ans 1ll * a * ans % mod;}a 1ll * a * a % mod;n 1;}return ans; }void get_r(int lim) {for (int i 0; i lim; i) {r[i] (i 1) * (lim 1) (r[i 1] 1);} }void get_inv(int n) {inv[1] 1;for (int i 2; i n; i) {inv[i] 1ll * (mod - mod / i) * inv[mod % i] % mod;} }void NTT(int *f, int lim, int rev) {for (int i 0; i lim; i) {if (i r[i]) {swap(f[i], f[r[i]]);}}for (int mid 1; mid lim; mid 1) {int wn quick_pow(3, (mod - 1) / (mid 1));for (int len mid 1, cur 0; cur lim; cur len) {int w 1;for (int k 0; k mid; k, w 1ll * w * wn % mod) {int x f[cur k], y 1ll * w * f[cur mid k] % mod;f[cur k] (x y) % mod, f[cur mid k] (x - y mod) % mod;}}}if (rev -1) {int inv quick_pow(lim, mod - 2);reverse(f 1, f lim);for (int i 0; i lim; i) {f[i] 1ll * f[i] * inv % mod;}} }void polyinv(int *f, int *g, int n) {if (n 1) {g[0] quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n 1 1);for (int i 0; i n; i) {t[i] f[i];}int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i 0; i lim; i) {int cur (2 - 1ll * g[i] * t[i] % mod mod) % mod;g[i] 1ll * g[i] * cur % mod;t[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;} }void polysqrt(int *f, int *g, int n) {if (n 1) {g[0] Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n 1 1);polyinv(g, b, n);int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);for (int i 0; i n; i) {t[i] f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i 0; i lim; i) {g[i] (1ll * inv2 * g[i] % mod 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] t[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;} }void derivative(int *a, int *b, int n) {for (int i 0; i n; i) {b[i] 1ll * a[i 1] * (i 1) % mod;} }void integrate(int *a, int n) {for (int i n - 1; i 1; i--) {a[i] 1ll * a[i - 1] * inv[i] % mod;}a[0] 0; }void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i 0; i lim; i) {g[i] 1ll * g[i] * b[i] % mod;b[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;}integrate(g, n); }void polyexp(int *f, int *g, int n) {if (n 1) {g[0] 1;return ;}polyexp(f, g, n 1 1);int lim 1;while (lim 2 * n) {lim 1;}polyln(g, d, n);for (int i 0; i n; i) {t[i] (f[i] - d[i] mod) % mod;}t[0] (t[0] 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i 0; i lim; i) {g[i] 1ll * g[i] * t[i] % mod;t[i] d[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;} }/*b存放多项式逆c存放多项式开根d存放多项式对数lne存放多项式指数expt作为中间转移数组,如果要用到polyinv得提前调用get_inv(n)先预先得到我们想要得到的逆元范围。 */int n, m, T, sa[N], sb[N], f1[20][N], f2[20][N], fac[N], ifac[N];void solve(int *f, int l, int r, int cur) {if (l r) {f1[cur][0] 1, f1[cur][1] mod - f[l];return ;}int mid l r 1, len1 mid - l 1, len2 r - mid;solve(f, l, mid, cur 1);for (int i 0; i len1; i) {f2[cur 1][i] f1[cur 1][i];f1[cur 1][i] 0;}solve(f, mid 1, r, cur 1);int lim 1;while (lim r - l 1) {lim 1;}get_r(lim);NTT(f1[cur 1], lim, 1);NTT(f2[cur 1], lim, 1);for (int i 0; i lim; i) {f1[cur][i] 1ll * f1[cur 1][i] * f2[cur 1][i] % mod;f1[cur 1][i] f2[cur 1][i] 0;}NTT(f1[cur], lim, -1); }int main() {// freopen(in.txt, r, stdin);// freopen(out.txt, w, stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf(%d %d, n, m);for (int i 1; i n; i) {scanf(%d, sa[i]);}for (int i 1; i m; i) {scanf(%d, sb[i]);}scanf(%d, T);int maxn max({n, m, T});fac[0] ifac[0] 1;for (int i 1; i 2 * maxn; i) {fac[i] 1ll * fac[i - 1] * i % mod;}ifac[2 * maxn] quick_pow(fac[2 * maxn], mod - 2);for (int i 2 * maxn - 1; i 1; i--) {ifac[i] 1ll * ifac[i 1] * (i 1) % mod;}get_inv(4 * maxn);solve(sa, 1, maxn, 0);for (int i 0; i maxn; i) {sa[i] f1[0][i];f1[0][i] 0;}polyln(sa, d, maxn 1);for (int i 0; i maxn; i) {sa[i] mod - 1ll * i * d[i] % mod;sa[i] 1ll * sa[i] * ifac[i] % mod;d[i] 0;}sa[0] (sa[0] n) % mod;solve(sb, 1, maxn, 0);for (int i 0; i maxn; i) {sb[i] f1[0][i];f1[0][i] 0;}polyln(sb, d, maxn 1);for (int i 0; i maxn; i) {sb[i] mod - 1ll * i * d[i] % mod;sb[i] 1ll * sb[i] * ifac[i] % mod;d[i] 0;}sb[0] (sb[0] m) % mod;int lim 1;while (lim 2 * maxn) {lim 1;}get_r(lim);NTT(sa, lim, 1);NTT(sb, lim, 1);for (int i 0; i lim; i) {sa[i] 1ll * sa[i] * sb[i] % mod;}NTT(sa, lim, -1);int inv quick_pow(1ll * n * m % mod, mod - 2);for (int i 1; i T; i) {printf(%d\n, 1ll * inv * sa[i] % mod * fac[i] % mod);}return 0; }