随笔分类 -  数学——FFT、NTT、多项式

摘要:"ref" cpp include include using namespace std; typedef long long ll; int n, a[270005], b[270005], c[270005], rev[270005]; const int mod=998244353, gg= 阅读全文
posted @ 2018-06-07 17:12 poorpool 阅读(189) 评论(0) 推荐(0)
摘要:我发现我的构造方法好像不太一样而且比较显然?……先读入 $q$ 数组(下表从零开始)。 记 $i j$ 时,$a_{i j}=1/i^2$;$i = j$ 时,$a_{i j}=0$。 答案 $E_i=\sum_{j=0}^{n 1}a_{i j}q_j$,可以用 FFT 优化,于是就做完了……吗? 阅读全文
posted @ 2018-05-23 22:07 poorpool 阅读(111) 评论(0) 推荐(0)
摘要:大概就是一个形状要嵌到矩阵里头这样子,看看代码就明白了qwq cpp include include include include include using namespace std; int n, m, k, lim=1, limcnt, rev[8388655]; char ss[5000 阅读全文
posted @ 2018-05-23 09:16 poorpool 阅读(465) 评论(0) 推荐(0)
摘要:"推柿子" "第二类斯特林数的容斥表达" fft卡精度就用ntt吧qwq。 cpp include include using namespace std; typedef long long ll; int n, lim=1, limcnt, rev[300005], inv[300005], a 阅读全文
posted @ 2018-05-04 21:23 poorpool 阅读(108) 评论(0) 推荐(0)
摘要:所有的下标从 $0$ 开始。 考虑枚举 $C$ (第一个加上负的等于第二个加上其绝对值)和第二个手链的偏移量 $p$。答案就是 $$\sum_{i=0}^{n 1}(x_i+C y_{(i+p) \bmod n})^2$$ 复制一遍 $y$ 数组就能去掉取模了,再展开就是 $$\sum_{i=0}^ 阅读全文
posted @ 2018-04-13 16:08 poorpool 阅读(162) 评论(0) 推荐(0)
摘要:"there" cpp include include include include include include using namespace std; int n, m, aa[300005], bb[300005], lim=1, tmpcnt, rev[1100005]; const 阅读全文
posted @ 2018-04-10 11:57 poorpool 阅读(103) 评论(0) 推荐(0)
摘要:"这里" cpp include include include include include using namespace std; typedef long long ll; int T, n, lim, c[100005], rev[270005], tmpcnt, a[100005]; 阅读全文
posted @ 2018-04-10 09:53 poorpool 阅读(76) 评论(0) 推荐(0)
摘要:参考资料 "picks" "miskcoo" "menci" "胡小兔" "unname" "自为风月马前卒" 上面是FFT的,学完了就来看 "NTT" 吧 "原根" 例题:luogu3803 fft优化后模板 cpp include include include using namespace 阅读全文
posted @ 2018-04-09 16:19 poorpool 阅读(255) 评论(0) 推荐(0)