[快速数论变换 NTT]

先粘一个模板。这是求高精度乘法的

#include <bits/stdc++.h>
#define maxn 1010
using namespace std;char s[maxn];typedef long long ll;
ll A[maxn], B[maxn];const int md = 998244353, G = 3;int n;ll power_mod(ll a, ll b){ll ret = 1;while(b > 0){if(b & 1)ret = ret * a % md;b >>= 1;a = a * a % md;}return ret;
}void NTT(ll A[], int n, int type){for(int i = 0, j = 0; i < n; i ++){if(i < j)swap(A[i], A[j]);for(int t = n >> 1; (j ^= t) < t; t >>= 1);}for(int k = 2; k <= n; k <<= 1){ll wn = type > 0 ? power_mod(G, (md - 1) / k) : power_mod(G, (md - 1) - (md - 1) / k);for(int i = 0; i < n; i += k){ll w = 1;for(int j = 0; j < k >> 1; j ++){ll T = w * A[i+j+(k>>1)] % md;A[i+j+(k>>1)] = (A[i+j] - T) % md;A[i+j] = (A[i+j] + T) % md;w = w * wn % md;}}}if(type < 0){ll inv = power_mod(n, md - 2);for(int i = 0; i < n; i ++)A[i] = A[i] * inv % md;}
}int main(){freopen("mul.in", "r", stdin);freopen("mul.out", "w", stdout);scanf("%s", s);int n1 = strlen(s);for(int i = n1 - 1; ~i; i --)A[n1-i-1] = s[i] ^ 48;scanf("%s", s);int n2 = strlen(s);for(int i = n2 - 1; ~i; i --)B[n2-i-1] = s[i] ^ 48;for(n = 1; n <= n1 + n2; n <<= 1);NTT(A, n, 1), NTT(B, n, 1);for(int i = 0; i < n; i ++)A[i] = A[i] * B[i] % md;NTT(A, n, -1);for(int i = 0; i < n; i ++){A[i] = (A[i] + md) % md;A[i+1] += A[i] / 10, A[i] %= 10;}while(n && A[n] == 0) n --;for(int i = n; i >= 0; i --)printf("%lld", A[i]);return 0;
}

[SDOI 2015]序列统计

取指标变成加法问题，构造生成函数F(x)，F(x)的n次方就是答案。

注意指标要mod M-1

还有注意0的问题，如果x!=0就不用管数据中的0，否则需要两种情况求和(然而数据中x并不等于0我就没写)

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define maxn 50010
using namespace std;
typedef long long ll;
const int md = 1004535809, G = 3;int N, M, X, S, a[maxn], Log[maxn], n;
ll F[maxn], H[maxn], ret[maxn];int p[maxn], primes, vis[maxn];ll power_mod(ll a, ll b, ll md){ll ret = 1;while(b > 0){if(b & 1) ret = ret * a % md;b >>= 1;a = a * a % md;}return ret;
}void pre_prime(){int n = 20000;for(int i = 2; i <= n; i ++){if(!vis[i])p[++ primes] = i;for(int j = 1; j <= primes; j ++){if(i * p[j] > n)break;vis[i * p[j]] = true;if(i % p[j] == 0)break;}}
}int getG(int M){pre_prime();int ret = 2;for(; ; ret ++){bool flag = true;for(int i = 1; i <= primes; i ++){if(M-1 < p[i])break;if((M-1) % p[i] == 0){if(power_mod(ret, (M - 1) / p[i], M) == 1){flag = false;break;}}}if(flag)return ret;}
}void NTT(ll A[], int n, int type){for(int i = 0, j = 0; i < n; i ++){if(i > j)swap(A[i], A[j]);for(int t = n >> 1; (j ^= t) < t; t >>= 1);}for(int k = 2; k <= n; k <<= 1){ll wn = type > 0 ? power_mod(G, (md-1)/k, md) : power_mod(G, md-1-(md-1)/k, md);for(int i = 0; i < n; i += k){ll w = 1;for(int j = 0; j < k >> 1; j ++){ll T = w * A[i+j+(k>>1)];A[i+j+(k>>1)] = (A[i+j] - T + md) % md;A[i+j] = (A[i+j] + T) % md;w = w * wn % md;}}}if(type < 0){ll inv = power_mod(n, md - 2, md);for(int i = 0; i < n; i ++)A[i] = A[i] * inv % md;}
}void CF(ll A[]){for(int i = 0; i < n; i ++)H[i] = F[i];NTT(A, n, 1), NTT(H, n, 1);for(int i = 0; i < n; i ++)A[i] = A[i] * H[i] % md;NTT(A, n, -1);for(int i = (M - 1); i < n; i ++)(A[i % (M - 1)] += A[i]) %= md, A[i] = 0;
}void power(int b){ret[0] = 1;while(b){if(b & 1)CF(ret);b >>= 1;CF(F);}
}int main(){scanf("%d%d%d%d", &N, &M, &X, &S);for(int i = 1; i <= S; i ++)scanf("%d", &a[i]);for(n = 1; n <= M; n <<= 1); n <<= 1;ll nw = 1, Gn = getG(M); for(int i = 0; i < M-1; i ++)Log[nw] = i, nw = nw * Gn % M;for(int i = 1; i <= S; i ++)if(a[i])F[Log[a[i]]] = 1;power(N);printf("%lld\n", (ret[Log[X]] + md) % md);return 0;
}

[COGS 2287]疯狂的机器人

这里有Catalan数的链接

#include <bits/stdc++.h>
#define maxn 300010
using namespace std;
typedef long long ll;
int n;
const int md = 998244353, G = 3;ll power_mod(ll a, ll b = md - 2){ll ret = 1;while(b > 0){if(b & 1)ret = ret * a % md;b >>= 1;a = a * a % md;}return ret;
}ll fac[maxn], inv[maxn], f[maxn];void NTT(ll* A, int n, int type){for(int i = 0, j = 0; i < n; i ++){if(i > j)swap(A[i], A[j]);for(int t = n >> 1; (j ^= t) < t; t >>= 1);}for(int k = 2; k <= n; k <<= 1){ll wn = power_mod(G, type > 0 ? (md-1) / k : md-1 - (md-1) / k);for(int i = 0; i < n; i += k){ll w = 1;for(int j = 0; j < k >> 1; j ++){ll T = w * A[i+j+(k>>1)] % md;A[i+j+(k>>1)] = (A[i+j] - T) % md;A[i+j] = (A[i+j] + T) % md;w = w * wn % md;}}}if(type == -1){ll inv = power_mod(n);for(int i = 0; i < n; i ++)A[i] = A[i] * inv % md;}
}int main(){freopen("crazy_robot.in", "r", stdin);freopen("crazy_robot.out", "w", stdout);scanf("%d", &n);fac[0] = inv[0] = 1;for(int i = 1; i <= 2 * n; i ++)fac[i] = fac[i-1] * i % md;inv[2 * n] = power_mod(fac[2 * n]);for(int i = 2 * n - 1; i >= 1; i --)inv[i] = inv[i+1] * (i+1) % md;for(int i = 0; i <= n; i ++){if(i & 1){f[i] = 0;continue;}f[i] = fac[i] * inv[i / 2] % md * inv[i / 2 + 1] % md * inv[i] % md;}int _N = 1;for(; _N <= n + n; _N <<= 1);NTT(f, _N, 1);for(int i = 0; i < _N; i ++)f[i] = f[i] * f[i] % md;NTT(f, _N, -1);ll ans = 0;for(int i = 0; i <= n; i ++)(ans += inv[n - i] * f[i]) %= md;cout << (ans * fac[n] % md + md) % md << endl;return 0;
}

[NOI十连测3007]卡常大法好。

#include <cstring>
#include <cstdio>
#define maxn 150010const int md = 998244353, G = 3;int R;
#define fastcall __attribute__((optimize("-Os")))
#define IL __inline__ __attribute__((always_inline))
fastcall IL int mul_mod(int a, int b){__asm__ __volatile__ ("\tmull %%ebx\n\tdivl %%ecx\n" :"=d"(R):"a"(a),"b"(b),"c"(md));return R;
}int n, N;int f[maxn], g[maxn], h[maxn], fac[maxn], inv[maxn], pw[maxn], tmp[maxn];fastcall IL int power_mod(int a, long long b = md - 2){int ret = 1;while(b > 0){if(b & 1)ret = mul_mod(ret, a);b >>= 1;a = mul_mod(a, a);}return ret;
}fastcall IL void swap(int& a, int& b){a ^= b ^= a ^= b;
}fastcall IL void NTT(int* A, int n, int type){for(int i = 0, j = 0; i < n; i ++){if(i > j)swap(A[i], A[j]);for(int t = n >> 1; (j ^= t) < t; t >>= 1);}for(int k = 2; k <= n; k <<= 1){int wn = power_mod(G, type > 0 ? (md-1)/k : (md-1)-(md-1)/k);for(int i = 0, m = k >> 1; i < n; i += k){int w = 1;for(int j = 0; j < k >> 1; j ++){int T = mul_mod(w, A[i+j+m]);A[i+j+m] = (A[i+j] - T + md) % md;A[i+j] = (A[i+j] + T) % md;w = mul_mod(w, wn);}}}if(type < 0){int inv = power_mod(n);for(int i = 0; i < n; i ++)A[i] = mul_mod(A[i], inv);}
}fastcall void Getinv(int* a, int* b, int n){if(n == 1){b[0] = power_mod(a[0]); return;}Getinv(a, b, n >> 1);memcpy(tmp, a, sizeof(a[0]) * n);NTT(tmp, n << 1, 1);NTT(b, n << 1, 1);for(int i = 0; i < n << 1; i ++)b[i] = mul_mod(b[i], (2ll - mul_mod(b[i], tmp[i]) + md) % md);NTT(b, n << 1, -1);memset(b + n, 0, sizeof(a[0]) * n);
}int F[30010][16];
int K;fastcall void cdq(int l, int r){if(l == r)return;int mid = l + r >> 1, len = r - l + 1;cdq(l, mid);for(n = 1; n <= len; n <<= 1);for(int i = 0; i < n; i ++)f[i] = h[i] = 0;for(int i = l; i <= mid; i ++)f[i-l] = mul_mod(F[i][K-1] + F[i][K], inv[i]);h[0] = 0;for(int i = 1; i < n; i ++)h[i] = mul_mod(g[i], inv[i-1]);if(n<=32) {for(int i = 0; i < n; i ++)tmp[i] = 0;for(int i = 0 ; i < n ; ++ i) {for(int j = 0 ; j <= i ; ++ j) {tmp[i] += mul_mod(h[j], f[i-j]);if(tmp[i]>=md) tmp[i] -= md;}}for(int i = 0 ; i < n ; ++ i) h[i] = tmp[i];} else {NTT(f, n, 1), NTT(h, n, 1);for(int i = 0; i < n; i ++)h[i] = mul_mod(h[i], f[i]);NTT(h, n, -1);}for(int i = mid + 1; i <= r; i ++){F[i][K] += mul_mod(h[i-l], fac[i-1]);if(F[i][K] >= md)F[i][K] -= md;}cdq(mid + 1, r);
}int main(){N = 30000;for(n = 1; n <= N; n <<= 1);fac[0] = inv[0] = 1;for(int i = 1; i <= n; i ++)fac[i] = mul_mod(fac[i - 1], i);inv[n] = power_mod(fac[n]);for(int i = n - 1; i >= 1; i --)inv[i] = mul_mod(inv[i + 1], i + 1);for(long long i = 0; i <= n; i ++)pw[i] = power_mod(2, i * (i - 1) / 2);for(int i = 1; i < n; i ++)h[i] = mul_mod(pw[i], inv[i - 1]);for(int i = 0; i < n; i ++)f[i] = mul_mod(pw[i], inv[i]);Getinv(f, g, n);n <<= 1;NTT(g, n, 1);NTT(h, n, 1);for(int i = 0; i < n; i ++)g[i] = mul_mod(g[i], h[i]);NTT(g, n, -1);for(int i = 1; i <= N; i ++)g[i] = mul_mod(g[i], fac[i - 1]);for(int i = 0; i <= N; i ++)F[i][0] = pw[i];for(int i = 1; i <= 15; i ++)K = i, cdq(0, N);static int S[20][20];S[0][0] = 1;for(int i = 1; i <= 15; i ++)for(int j = 1; j <= i; j ++){S[i][j] = mul_mod(S[i-1][j], j) + S[i-1][j-1];if(S[i][j] >= md)S[i][j] -= md;}int test, n, m;scanf("%d", &test);while(test --){scanf("%d%d", &n, &m);int ans = 0;for(int i = 0; i <= m; i ++){ans += mul_mod(mul_mod(S[m][i], F[n][i]), fac[i]);if(ans >= md)ans -= md;}printf("%d\n", ans);}return 0;
}

转载于:https://www.cnblogs.com/Candyouth/p/5437789.html

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