# QOJ

#Nom d'utilisateurMottoRésolu
/Qingyu5829
1larryzhong926
2xiaowuc1you're a half a world away, but in my mind I whisper every single word you say517
3MaMengQi459
4ZhaoZiLong456
5HuangHanSheng439
6ZhangYiDe435
7GuanYunchang432
/flowersqytxdy!325
8hyddQingyu txdy $\\$ $\text{If my armor breaks, I'll fuse it back together}$312
9Crysfly$$f(x)=(\sum_{i=0}^{n-1}\frac{y_i}{(x-q^i)\prod_{j\ne i}(q^i-q^j)})\prod_{i=0}^{n-1}(x-q^i)$$ 247
10tricyzhkx228
11chenshi215
12zhouhuanyi199
13Wu_Ren189
14maspy187
15repoman$$\prod_{i=0}^{n-1} (1+q^iz) = \sum_{i=0}^n q^{i(i-1)/2}\binom ni_q z^i$$176
16He_Ren170
16qwq$\displaystyle \sum_{i=1}^n [i,i+1,\cdots, i+k] \pmod{10^9+7}$170
18alpha1022$$\frac{1}{n_1!n_2!}(1-y)^{n_1+n_2+2} \left(\sum_{j\ge 0} y^j(t+j)^{n_1} \right) \left(\sum_{j\ge 0} y^j((j+1)-t)^{n_2} \right)$$163
19ckiseki161
20feecle6418gyh ak ioi160
21Lenstarorz Qingyu156
21Sa3tElSefrleh156
23Appleblue17152
24zhangboju短暂登上首页并即将掉下来149
25eyiigjkn147
26BeyondHeaven138
26hutality$$[x^n](1-x)^{-\frac{1}{2}} \exp\left(\frac{x(1+x)}{2-2x}\right)\frac{\left(\frac{2-x}{2-2x}\right)^k}{k!}\cdot\,_0 F _1 \left(;k+1;x\left(\frac{2-x}{2-2x}\right)^2\right)$$138
28lmeowdn137
29LH$$|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^g|$$135
29Minneapolis卷王别卷了135
31YaoBIG131
32zombie462<b>123</b>124
33Sorting123
34not_so_organic今晚九点，QYC 唱歌。121
35JohnAlfnovQingyu Kedavra ! 119
36whatever116
37std大家好，我是来自彭博社的埃斯提迪115
38sdoi$$P_{a_i}(x) = [t^{a_i}]\frac{1}{1-xF(t)}$$111
39zhoukangyang110
40DianasDog关注嘉然，顿顿解馋109
41AFewSuns$\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$108
42hos_lyric105
/Qiuly理论上状态数应该是 $2^{552}$，但搜一搜发现只有 $1834$ 。104
43xiaoyaowudi103
44propensityThe 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
45ybw051114100
46LoverInTime089年的树剖99
47myee与其诺诺以顺，不若谔谔以昌98
47smax98
47wlxhkk98
50perspective97
51Baltinic$$\rho \left ( \frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v} \right) = - \nabla p + \nabla \cdot \mathbf{T} + \mathbf{f}$$92
51DaBenZhongXiaSongKuaiDi92
53George_Plover91
55huredomirikayatuda90
56testusera88
57little_sun86
58yangjiuzhi@_silhouette_84
59Beevo83
60abdelrahman00182
6115867449938$$Z_G(t_1,t_2,\ldots,t_n) = \frac{1}{|G|}\sum_{g \in G} t_1^{c_1(g)} t_2^{c_2(g)} \cdots t_n^{c_n(g)}$$81
6199pts_WA81
63123456俺宣布俺是正宗滴【123456】79
64Chenguanlin陈冠霖培训班现在开课，报名请私聊陈冠霖。76
64_silhouette_76
66fzj200775
66Macesuted75
68PetroTarnavskyi72
68xiaojifang大家好我是小机房，又名更衣室。72
70Froggygua周转没有丁丁。71
70hydd_lenstar_team71
72winmain70
73triplem5ds69
74sagittarius_fjz魔法少女小熊68
75Mr_Eight今晚九点，whq唱歌，不见不散。67
76goodman66
77KING_UT65
77pretentious$$\det(AB) = \sum_{S\in\tbinom{[n]}m} \det(A_{[m],S})\det(B_{S,[m]})$$65
79neko_nyaa63
79SuffixTree63
81Arraiter62
81CharlieVinnieYou fly away so proudly, away from my summer. Slient words of praying, all these years repeating62
81littlesummer62
81MIT0162
81znstz62
86flywatre61
86skittles141261
86taniya$$\prod_{n=1}^\infty (1-x^n)=\sum_{k=-\infty}^\infty(-1)^kx^{\frac{k(3k-1)}{2}}=\sum_{k=0}^\infty(-1)^kx^{\frac{k(3k\pm 1)}{2}}$$61
8900$$e^x = \sum_{n=0} \frac{x^n}{n!}$$60
89qinjianbin60
91275307894a59
91Iris?59
91QAQQWQ59
94magicduck57
94Zuqa57
96JackF______是不是贺子56
96yh56
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