QOJ.ac

QOJ

#Nom d'utilisateurMottoRésolu
1xiaowuc1you're a half a world away, but in my mind I whisper every single word you say1307
2hos_lyric$$\left[\frac{x^n}{n!} q^{K-1}\right] \frac{(\mathrm{e}^{rx} - 1) (q + r + \mathrm{e}^{rx} + q \mathrm{e}^{rx} - r \mathrm{e}^{rx} + 1)}{(q + 1) (r - \mathrm{e}^{2rx} + r \mathrm{e}^{2rx} + 1)}$$1160
3ucup-team1005沉舟侧畔千帆过1035
4larryzhong983
5PetroTarnavskyiWhy haven't you registered at https://algotester.com/en yet? 🤨798
6Crysfly我还能为你驻足到何时672
7PhantomThreshold669
8zhouhuanyi664
9ckiseki卷不动了608
10ucup-team2335563
11dXqwqTo the cosmic547
12qiuzx546
13ucup-team004532
14lmq26052003531
15grass8cow521
16ucup-team1198515
17ucup-team087490
18SolitaryDream462
19ucup-team052458
20ucup-team1209我不是可爱小青鱼442
21ucup-team2880431
22ucup-team1878418
23maspyUniversal Cup Upsol部405
24LoverInTime089年的树剖401
25275307894a394
26flower377
27ucup-team1004方队akioi373
28hyforces永远不够366
29ucup-team987355
30Kevin5307$$ \frac{1}{1-x}\sum_{k=1}^\infty\frac{x^k}{1-\frac{x-x^{k+1}}{1-x}} $$353
31bachbeo2007349
31ship2077349
33ucup-team159346
34ucup-team191343
35ucup-team864340
36MoRanSky无论过去 不问将来337
37ucup-team1134326
38lmeowdn316
38waifuSenpai316
40ucup-team133314
41hyddQingyu txdy $\\$ If my armor breaks, I'll fuse it back together310
42ucup-team130301
43Lynkcat299
43Sortingㅗ오ㅗ 299
43ucup-team253299
46ucup-team266290
47james1BadCreeper他明白 他明白 我给不起287
48alpha1022$$\frac{1}{n_1!n_2!}(1-y)^{n_1+n_2+2} \left(\sum_{j\ge 0} y^j(t+j)^{n_1} \right) \left(\sum_{j\ge 0} y^j((j+1)-t)^{n_2} \right)$$286
49ucup-team138283
50ucup-team112282
51znstz278
52ucup-team025274
53AFewSuns$\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$269
54ezteam1267
55ucup-team1525266
56ucup-team1447没队要262
57ucup-team180257
58zhaohaikun256
59tricyzhkx250
60ucup-team216248
60ucup-team311248
62ucup-team2307247
63KKT89245
64ucup-team267哈姆。242
64zhoukangyang242
66ucup-team244241
67iee238
68He_Ren237
68sdoi$$P_{a_i}(x) = [t^{a_i}]\frac{1}{1-xF(t)}$$237
70ucup-team122236
71ucup-team134235
728BQube222
73chenshi221
73karuna221
75ucup-team3215217
76repoman$$\prod_{i=0}^{n-1} (1+q^iz) = \sum_{i=0}^n q^{i(i-1)/2}\binom ni_q z^i$$215
76ucup-team870215
76ushg8877215
79ucup-team1293214
80ucup-team1123212
80ucup-team228212
82JohnAlfnovQingyu Kedavra ! 210
83ucup-team2818208
84chenxinyang2006206
85do_while_true204
86new_dawn_2203
86ucup-team045203
88kevinyang201
88ucup-team1002201
88ucup-team123201
91DaiRuiChen007199
91Max_s_xaM199
91qwq$\displaystyle \sum_{i=1}^n [i,i+1,\cdots, i+k] \pmod{10^9+7}$199
91ucup-team1126199
95ucup-team029197
96ucup-team1191196
97Wu_Ren194
98rotcar08193
98ucup-team635193
100PlentyOfPenalty高亮的是我们,我们是青鱼家军!192
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