乒乓球牛客 NTT

乒乓球

solution

注意：以下k=i−j+1表示i,j中间元素的个数，i<j注意：以下~k=i-j+1表示i,j中间元素的个数，i<j注意：以下 k=i−j+1表示i,j中间元素的个数，i<j

考虑拿出顺序，对于i,j来说，它们之间有k个数(0<k<n−1),因此有2!∗k!种，剩下n−k−1个球插空放入(k+3)(k+3)...(n−1)n，且总方案为n!，易得：考虑拿出顺序，对于i,j来说，它们之间有k个数(0<k<n-1),因此有2!*k!种，剩下n-k-1个球插空放入(k+3)(k+3)...(n-1)n，且总方案为n!，易得：考虑拿出顺序，对于i,j来说，它们之间有k个数(0<k<n−1),因此有2!∗k!种，剩下n−k−1个球插空放入(k+3)(k+3)...(n−1)n，且总方案为n!，易得：

∑i=1n−2∑j=3n2!∗k!∗(k+3)(k+3)...(n−1)n∗(wiwj)n!\frac{\sum\limits_{i=1}^{n-2}\sum\limits_{j=3}^{n}2!*k!*(k+3)(k+3)...(n-1)n*(w_iw_j)}{n!}n!i=1∑n−2j=3∑n2!∗k!∗(k+3)(k+3)...(n−1)n∗(wiwj)

=∑k=1n−2∑j=3nwiwj2n!(k+1)(k+2)n!=\frac{\sum\limits_{k=1}^{n-2}\sum\limits_{j=3}^{n}w_iw_j\frac{2n!}{(k+1)(k+2)}}{n!}=n!k=1∑n−2j=3∑nwiwj(k+1)(k+2)2n!

=2∑k=1n−2∑j=3nwiwj(k+1)(k+2)=2\sum\limits_{k=1}^{n-2}\sum\limits_{j=3}^{n}\frac{w_iw_j}{(k+1)(k+2)}=2k=1∑n−2j=3∑n(k+1)(k+2)wiwj

=2∑j=3nwj∑k=1n−2wj−k−1(k+1)(k+2)=2\sum\limits_{j=3}^{n}w_j\sum\limits_{k=1}^{n-2}{\frac{w_{j-k-1}}{(k+1)(k+2)}}=2j=3∑nwjk=1∑n−2(k+1)(k+2)wj−k−1

显然对∑k=1n−2wj−k−1∗1(k+1)(k+2)进行卷积之后分别与wj作积再累加即可.显然对\sum\limits_{k=1}^{n-2}{w_{j-k-1}*\frac{1}{(k+1)(k+2)}}进行卷积之后分别与w_j作积再累加即可.显然对k=1∑n−2wj−k−1∗(k+1)(k+2)1进行卷积之后分别与wj作积再累加即可.

code

/*SiberianSquirrel*//*CuteKiloFish*/
#include <bits/stdc++.h>
using namespace std;
#define gcd(a,b) __gcd(a,b)
#define Polynomial vector<int>
#define Inv(x) quick_pow(x, mod - 2)
using ll = long long;
using ull = unsigned long long;
const ll mod = 998244353, mod_g = 3, img = 86583718;//NTT, 多项式三角形
const int N = int(6e5 + 10);inline int add(int a, int b) {return a + b < mod? a + b: a + b - mod;
}
inline int sub(int a, int b) {return a - b < 0? a - b + mod: a - b;
}
inline int mul(int a, int b) {return 1ll * a * b % mod;
}void print(Polynomial &a, int len){for(int i = 0; i < len; ++ i)cout << a[i] << ' ';cout << '\n';
}
void exgcd(int a,int b,int &x,int &y){if(!b) {x = 1; y = 0;return;}exgcd(b, a % b, y, x);y -= x * (a / b);
}
int quick_pow(int ans, int p, int res = 1) {for(; p; p >>= 1, ans = 1ll * ans * ans % mod)if(p & 1) res = 1ll * res * ans % mod;return res % mod;
}Polynomial R;
//二进制向上取整，为方便NTT变换准备。
inline int Binary_Rounding(const int &n) {int len = 1; while(len < n) len <<= 1;return len;
}
//预处理R数组，准备变换，在每次NTT之前理论都要调用此函数。
inline int Prepare_Transformation(int n){int L = 0, len;for(len = 1; len < n; len <<= 1) L++;R.clear(); R.resize(len);for(int i = 0; i < len; ++ i)R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));return len;
}
void NTT(Polynomial &a, int f){int n = a.size();for(int i = 0; i < n; ++ i)if(i < R[i])swap(a[i], a[R[i]]);for(int i = 1; i < n; i <<= 1)for(int j = 0, gn = quick_pow(mod_g,(mod - 1) / (i<<1)); j < n; j += (i<<1))for(int k = 0, g = 1, x, y; k < i; ++ k, g = 1ll * g * gn % mod)x = a[j + k], y = 1ll * g * a[i + j + k] % mod,a[j + k] = (x + y) % mod, a[i + j + k] = (x - y + mod) % mod;if(f == -1){reverse(a.begin() + 1, a.end());int inv = Inv(n);for(int i = 0; i < n; ++ i) a[i] = 1ll * a[i] * inv % mod;}
}inline Polynomial operator +(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i)ret[i] = (1ll * a[i] + b + mod) % mod;return ret;
}
inline Polynomial operator -(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i)ret[i] = (1ll * a[i] - b + mod) % mod;return ret;
}
inline Polynomial operator *(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i) ret[i] = (1ll * a[i] * b % mod + mod) % mod;return ret;
}
inline Polynomial operator +(const Polynomial &a, const Polynomial &b){int sizea = a.size(), sizeb = b.size(), size = max(sizea, sizeb);Polynomial ret = a; ret.resize(size);for(int i = 0; i < sizeb; ++ i) ret[i] = (1ll * ret[i] + b[i]) % mod;return ret;
}
inline Polynomial operator -(const Polynomial &a, const Polynomial &b){int sizea = a.size(), sizeb = b.size(), size = max(sizea, sizeb);Polynomial ret = a; ret.resize(size);for(int i = 0; i < sizeb; ++ i) ret[i] = (1ll * ret[i] - b[i] + mod) % mod;return ret;
}
inline Polynomial Inverse(const Polynomial &a){Polynomial ret, inv_a;ret.resize(1);ret[0] = Inv(a[0]); int ed = a.size();for(int len = 2; len <= ed; len <<= 1){int n = Prepare_Transformation(len << 1);inv_a = a; inv_a.resize(n); ret.resize(n);for(int i = len; i < n; ++ i) inv_a[i] = 0;NTT(inv_a, 1); NTT(ret, 1);for(int i = 0; i < n; ++ i)ret[i] = 1ll * (2ll - 1ll * inv_a[i] * ret[i] % mod + mod) % mod * ret[i] % mod;NTT(ret, -1);for(int i = len; i < n; ++ i) ret[i] = 0;}ret.resize(ed);return ret;
}
inline Polynomial operator *(const Polynomial &a, const Polynomial &b){Polynomial lsa = a, lsb = b, ret;int n = lsa.size(), m = lsb.size();n = Prepare_Transformation(n + m);lsa.resize(n); lsb.resize(n); ret.resize(n);NTT(lsa,1); NTT(lsb,1);for(int i = 0; i < n; ++ i) ret[i] = 1ll * lsa[i] * lsb[i] % mod;NTT(ret,-1);return ret;
}
inline Polynomial operator /(const Polynomial &a, const Polynomial &b){Polynomial ret = a, ls = b;reverse(ret.begin(), ret.end());reverse(ls.begin(), ls.end());ls.resize(Binary_Rounding(a.size() + b.size()));ls = Inverse(ls);ls.resize(a.size() + b.size());ret = ret * ls; ret.resize(a.size() - b.size() + 1);reverse(ret.begin(), ret.end());return ret;
}
inline Polynomial operator %(const Polynomial &a, const Polynomial &b){Polynomial ret = a / b;ret = ret * b; ret.resize(a.size() + b.size());ret = a - ret; ret.resize(a.size() + b.size());return ret;
}
inline Polynomial Derivation(const Polynomial &a){int size = a.size(); Polynomial ret; ret.resize(size);for(int i = 1; i < size; ++ i) ret[i - 1] = 1ll * i * a[i] % mod;ret[size - 1] = 0;return ret;
}
inline Polynomial Integral(const Polynomial &a){int size = a.size(); Polynomial ret; ret.resize(size);for(int i = 1; i < size; ++ i) ret[i] = 1ll * Inv(i) * a[i - 1] % mod;ret[0] = 0;return ret;
}
inline Polynomial Composition_Inverse(const Polynomial &a){int n = a.size();Polynomial ret, Cinv = a, Pow;Cinv.resize(n); ret.resize(n); Pow.resize(n); Pow[0] = 1;for(int i = 0; i < n - 1; ++ i) Cinv[i] = Cinv[i + 1]; Cinv[n - 1] = 0;Cinv = Inverse(Cinv);for(int i = 1; i < n; ++ i){Pow = Pow * Cinv; Pow.resize(n);ret[i] = 1ll * Pow[i - 1] * Inv(i) % mod;}return ret;
}
inline Polynomial Logarithmic(const Polynomial &a){Polynomial ln_a = Derivation(a) * Inverse(a);ln_a.resize(a.size());return Integral(ln_a);
}
inline Polynomial Exponential(const Polynomial &a, int Constant = 1){Polynomial ret, D; int ed = a.size();ret.resize(1); ret[0] = Constant;for(int len = 2; len <= ed; len <<= 1){D = Logarithmic(ret); D.resize(len);D[0] = (1ll * a[0] + 1ll - D[0] + mod) % mod;for(int i = 1; i < len; ++i) D[i] = (1ll * a[i] - D[i] + mod) % mod;int n = Prepare_Transformation(len<<1);ret.resize(n); D.resize(n);NTT(ret, 1); NTT(D,1);for(int i = 0; i < n; ++ i) ret[i] = 1ll * ret[i] * D[i] % mod;NTT(ret, -1);for(int i = len; i < (len<<1); ++ i) ret[i] = D[i] = 0;}ret.resize(ed);return ret;
}int fac[int(1e5 + 10)] = {1}, ifac[int(1e5 + 10)] = {1};inline void solve() {int n; cin >> n; vector<int> w(n + 1); for(int i = 1; i <= n; ++ i) cin >> w[i];Polynomial F(n + 1), G(n + 1);for(int i = 1; i < n - 1; ++ i) {F[i] = 1ll * Inv(i + 1) * Inv(i + 2) % mod;}for(int i = 1; i < n - 1; ++ i) G[n - i] = w[n - i - 1];Polynomial H = G * F;ll res = 0;for(int i = 3; i <= n; ++ i) {res = (res + H[i] * w[i] % mod) % mod;}cout << 2ll * res % mod << '\n';
}signed main() {ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#ifdef ACM_LOCALfreopen("input", "r", stdin);
//    freopen("output", "w", stdout);signed test_index_for_debug = 1;char acm_local_for_debug = 0;do {if (acm_local_for_debug == '$') exit(0);if (test_index_for_debug > 20)throw runtime_error("Check the stdin!!!");auto start_clock_for_debug = clock();solve();auto end_clock_for_debug = clock();cout << "Test " << test_index_for_debug << " successful" << endl;cerr << "Test " << test_index_for_debug++ << " Run Time: "<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;cout << "--------------------------------------------------" << endl;} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#elsesolve();
#endifreturn 0;
}