hdu 6061 RXD and functions [快速数论变换]

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题意 :

分析: 很容易可以得出:

即 :

最后可得每一项的系数为

然后可以用NTT(快速数论变换)求出每一项的系数.

#include<bits/stdc++.h>using namespace std;typedef long long ll;const ll mod = 998244353;const ll maxn = 1e5 + 10;ll wn[22];ll a[maxn << 2], b[maxn << 2];ll F[maxn], N[maxn];ll qmod(ll a, ll b) {    ll ans = 1;    while(b) {        if(b & 1) ans = ans * a % mod;        a = a * a % mod;        b >>= 1;    }    return ans;}void init() {    for(int i = 0; i < 20; i++) {        int t = 1 << i;        wn[i] = qmod(3, (mod - 1) / t);    }    F[0] = 1, N[0] = 1;    for(int i = 1; i < maxn; i++) {        F[i] = F[i - 1] * i % mod;    }    N[maxn - 1] = qmod(F[maxn - 1], mod - 2);    for(int i = maxn - 1; i >= 1; i--) {        N[i - 1] = N[i] * i % mod;    }}void bit_reverse(int n, ll *x) {    for(int i = 0, j = 0; i != n; i++) {        if(i > j) swap(x[i], x[j]);        for(int l = n >> 1; (j ^= l) < l; l >>= 1);    }}void NTT(ll *a, int len, int on) {    bit_reverse(len, a);    int id = 0;    for(int h = 2; h <= len; h <<= 1) {        id++;        for(int j = 0; j < len; j += h) {            ll w = 1;            for(int k = j; k < j + h / 2; k++) {                ll u = a[k] % mod;                ll t = w * (a[k + h / 2] % mod) % mod;                a[k] = (u + t) % mod;                a[k + h / 2] = ((u - t) % mod + mod) % mod;                w = w * wn[id] % mod;            }        }    }    if(on == -1) {        for(int i = 1; i < len / 2; i++)            swap(a[i], a[len - i]);        ll Inv = qmod(len, mod - 2);        for(int i = 0; i < len; i++)            a[i] = a[i] % mod * Inv % mod;    }}void Conv(ll *a, ll *b, int n) {    NTT(a, n, 1);    NTT(b, n, 1);    for(int i = 0; i < n; i++)        a[i] = a[i] * b[i] % mod;    NTT(a, n, -1);}int main() {    init();    int n, m;    while(~scanf("%d", &n)) {        memset(a, 0, sizeof a);        memset(b, 0, sizeof b);        for(int i = 0; i <= n; i++) {            scanf("%lld", &a[i]);            a[i] = a[i] * F[i] % mod;        }        scanf("%d", &m);        int s = 0, x;        for(int i = 0; i < m; i++) {            scanf("%d", &x);            s = (s + x) % mod;        }        s = (-s + mod) % mod;        ll g = 1;        for(int i = 0; i <= n; i++) {            b[n - i] = g * N[i] % mod;            g = g * s % mod;        }        int p = 1;        while(p <= n) p <<= 1;        p <<= 1;        Conv(a, b, p);        for(int i = 0; i <= n; i++) {            a[i + n] = a[i + n] * N[i] % mod;            a[i + n] = (a[i + n] + mod) % mod;            printf("%lld ", a[i + n]);        }        printf("\n");    }    return 0;}

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