QOJ.ac

QOJ

#UsernameMottoSolved
/Qingyu永眠6589
1larryzhong982
2MaMengQi708
3ZhangYiDe703
4ZhaoZiLong696
5HuangHanSheng685
6GuanYunchang681
7xiaowuc1you're a half a world away, but in my mind I whisper every single word you say527
8hos_lyric$$\left[\frac{x^n}{n!} q^{K-1}\right] \frac{(\mathrm{e}^{rx} - 1) (q + r + \mathrm{e}^{rx} + q \mathrm{e}^{rx} - r \mathrm{e}^{rx} + 1)}{(q + 1) (r - \mathrm{e}^{2rx} + r \mathrm{e}^{2rx} + 1)}$$465
9flowerqingyu txdy!357
10Crysflysunflower-addicted349
11hyddQingyu txdy $\\$ $\text{If my armor breaks, I'll fuse it back together}$309
12zhouhuanyi300
13maspy252
14PetroTarnavskyi251
15tricyzhkx248
16He_Ren237
17sdoi$$P_{a_i}(x) = [t^{a_i}]\frac{1}{1-xF(t)}$$234
18ckiseki229
19chenshi218
20lmeowdn213
21Sortingㅗ오ㅗ 212
22ucup-team004210
23repoman$$\prod_{i=0}^{n-1} (1+q^iz) = \sum_{i=0}^n q^{i(i-1)/2}\binom ni_q z^i$$208
24PhantomThreshold204
25qwq$\displaystyle \sum_{i=1}^n [i,i+1,\cdots, i+k] \pmod{10^9+7}$198
26Wu_Ren192
27AFewSuns$\displaystyle\sum_{n \geq 0}{(x+y)^n\frac{u^n}{n!}}=\sum_{k \geq 0}\frac{x}{x-kz}(x-kz)^k\frac{u^k}{k!}\sum_{l \geq 0}\frac{(y+kz)^l}{l!}u^l$171
/alpha1022$$\frac{1}{n_1!n_2!}(1-y)^{n_1+n_2+2} \left(\sum_{j\ge 0} y^j(t+j)^{n_1} \right) \left(\sum_{j\ge 0} y^j((j+1)-t)^{n_2} \right)$$171
27myee与其诺诺以顺,不若谔谔以昌171
29eyiigjkn170
29feecle6418gyh ak ioi170
31lmq26052003159
32Sa3tElSefrleh157
33Lenstarorz Qingyu154
34Appleblue17152
35zhangboju短暂登上首页并即将掉下来149
36BeyondHeaven143
37zhoukangyang139
38hutality$$[x^n](1-x)^{-\frac{1}{2}} \exp\left(\frac{x(1+x)}{2-2x}\right)\frac{\left(\frac{2-x}{2-2x}\right)^k}{k!}\cdot\,_0 F _1 \left(;k+1;x\left(\frac{2-x}{2-2x}\right)^2\right)$$138
39LH$$|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^g|$$136
40Minneapolis卷王别卷了135
/Qiulyfendou132
41YaoBIG131
42SolitaryDream128
43wlxhkk126
44JohnAlfnovQingyu Kedavra ! 124
44LoverInTime089年的树剖124
44qzez124
4799pts_WA123
47ucup-team1209123
47zombie462<b>123</b>123
50not_so_organic今晚九点,QYC 唱歌。121
51275307894a118
51whatever118
53znstz116
54ybw051114115
55DianasDog关注嘉然,顿顿解馋113
55new_dawn_2113
57neko_nyaa112
58triplem5ds111
59xiaoyaowudi110
60Beevo107
61bulijiojiodibuliduo104
62smax102
62waifuSenpai102
648BQube101
64propensityThe binary Gray code is fun, For in it strange things can be done. Fifteen, as you know, Is one, oh, oh, oh, And ten is one, one, one and one101
64UrgantTeam101
67sinbad100
68lpf98
69perspective$$\sqrt{\pi n} \lesssim \mathbb E[\mathrm{depth}] \leq O((n\log n)^{1/2})$$97
/yzhang如果结果不如你所愿,就在尘埃落定前奋力一搏96
70Baltinic$$\rho \left ( \frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v} \right) = - \nabla p + \nabla \cdot \mathbf{T} + \mathbf{f}$$93
71DaBenZhongXiaSongKuaiDi92
72DitaMirika91
72George_Plover91
72SorahISA91
75huredomirikayatuda90
76bashkortLost time is never found again.89
76chenchi89
76hyforces89
76rgnerdplayer89
76Scintilla89
81testusera88
82little_sun86
82xaphoenix86
84youngsystem84
85yangjiuzhi@_silhouette_83
86abdelrahman00182
86Forever_Young82
8815867449938$$Z_G(t_1,t_2,\ldots,t_n) = \frac{1}{|G|}\sum_{g \in G} t_1^{c_1(g)} t_2^{c_2(g)} \cdots t_n^{c_n(g)}$$81
88yyyyxh81
90123456俺宣布俺是正宗滴【123456】80
91fzj200778
91karuna78
93Chenguanlin陈冠霖培训班现在开课,报名请私聊陈冠霖。76
93LFCode无限拓展的玛瑙星空之下,少女伫立着倾听风的声音。76
95Macesuted75
96batrr74
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11