博主最近心血來潮，貼貼全家桶。
本多項式全家桶包含了：多項式乘法，多項式求逆，多項式求ln，多項式求exp，多項式除法，多項式取模，多項式多點求值，以及乙個為了多點求值用的分治ntt。
由於時代變了，博主的多點求值使用了常數較小的版本。
#include #define i inline
#define fi first
#define se second
#define ll long long
#define mp make_pair
#define pii pair#define fo(i, a, b) for(int i = a; i <= b; i++)
#define fd(i, a, b) for(int i = a; i >= b; i--)
#define ull unsigned long long
using namespace std;
const int inf = 2147483647;
const int mod = 998244353;
const int n = 1 << 17;
i int _max(int x, int y)
i int _min(int x, int y)
i ll read()
while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
return x * f;
}i void ptt(ll x)
i void put(ll x)
i void pr1(ll x)
i void pr2(ll x)
i int pow_mod(int a, int k)
namespace poly return len;
} i int pre(int n)
i int gao(int x)
i void dft(int y, int len)
i void idft(int y, int len)
i void clear(int y, int len)
i void clear(int a, int s, int t)
i void cpy(int a, int b, int len)
i void cpy(int a, vectorb)
i void cpy(vector&a, int b, int len)
i void getinv(int a, int b, int len)
getinv(a, b, len >> 1); cpy(h, a, len), clear(b, len);
fo(i, 0, (len << 1) - 1) r[i] = (r[i >> 1] >> 1) | ((i & 1) ? len : 0);
dft(h, len << 1), dft(b, len << 1);
fo(i, 0, (len << 1) - 1) b[i] = (2 - (ll)b[i] * h[i]) % mod * b[i] % mod;
idft(b, len << 1); clear(b, len);
} i void getln(int a, int b, int len)
const int z = 16;
int mem1[n * 10], z[n << 1]; ull mem2[n * 10], g[n << 1];
i void solve(int l, int r, int *memp1, ull *memp2) return ;
} int len = (r - l + 1) / z, ll = len << 1;
fo(i, 0, ll - 1) r[i] = (r[i >> 1] >> 1) | ((i & 1) ? (ll >> 1) : 0);
int *h1[z]; ull *h2[z];
fo(i, 0, z - 1) if(l == 0)
} fo(i, 0, z - 1)
} i void getexp(int a, int b, int len)
int c[n << 1];
i void getpw(int a, int b, int k, int len)
int v[n << 1];
i void mul(int a, int b, int len1, int len2)
cpy(v, b, len2), clear(a, len1, len), clear(v, len2, len);
dft(a, len), dft(v, len);
fo(i, 0, len - 1) a[i] = (ll)a[i] * v[i] % mod;
idft(a, len); clear(a, len1 + len2 - 1, len);
} i void mult(int a, int b, int len1, int len2) cpy(a, v, len1 - len2 + 1), clear(a, len1 - len2 + 1, len);
return ;
} cpy(v, b, len2), reverse(v, v + len2), clear(v, len2, len), clear(a, len1, len);
dft(a, len), dft(v, len);
fo(i, 0, len - 1) a[i] = (ll)a[i] * v[i] % mod;
idft(a, len); fo(i, 0, len1 - len2) a[i] = a[i + len2 - 1]; clear(a, len1 - len2 + 1, len);
} int h1[n << 1], h2[n << 1];
i void getdiv(int a, int b, int c, int n, int m)
i void getmod(int a, int b, int c, int n, int m)
vectorg[n << 1], f[n << 1]; int d[n], a[n << 1], b[n << 1];
i void gao1(int u, int l, int r) int mid = l + r >> 1;
gao1(u << 1, l, mid), gao1(u << 1 | 1, mid + 1, r);
cpy(a, g[u << 1]), cpy(b, g[u << 1 | 1]);
mul(a, b, mid - l + 2, r - mid + 1);
cpy(g[u], a, r - l + 2);
} i void gao2(int u, int l, int r)
int mid = l + r >> 1;
cpy(a, f[u]), cpy(b, g[u << 1 | 1]);
mult(a, b, r - l + 2, r - mid + 1);
cpy(f[u << 1], a, mid - l + 2);
cpy(a, f[u]), cpy(b, g[u << 1]);
mult(a, b, r - l + 2, mid - l + 2);
cpy(f[u << 1 | 1], a, r - mid + 1);
gao2(u << 1, l, mid), gao2(u << 1 | 1, mid + 1, r);
} i void getval(int a, int b, int n, int m)
}int main()
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