/ | Qingyu | | 5885 |
1 | larryzhong | | 931 |
2 | xiaowuc1 | you're a half a world away, but in my mind I whisper every single word you say | 519 |
3 | MaMengQi | | 462 |
4 | ZhaoZiLong | | 459 |
5 | HuangHanSheng | | 442 |
6 | ZhangYiDe | | 440 |
7 | GuanYunchang | | 437 |
/ | flower | qingyu txdy! | 334 |
8 | hydd | Qingyu txdy $\\$ $\text{If my armor breaks, I'll fuse it back together}$ | 312 |
9 | Crysfly | $$f(x)=(\sum_{i=0}^{n-1}\frac{y_i}{(x-q^i)\prod_{j\ne i}(q^i-q^j)})\prod_{i=0}^{n-1}(x-q^i)$$
| 254 |
10 | tricyzhkx | | 230 |
11 | chenshi | | 216 |
12 | zhouhuanyi | | 199 |
13 | maspy | | 193 |
14 | Wu_Ren | | 189 |
15 | He_Ren | | 183 |
16 | repoman | $$\prod_{i=0}^{n-1} (1+q^iz) = \sum_{i=0}^n q^{i(i-1)/2}\binom ni_q z^i$$ | 176 |
17 | qwq | $\displaystyle \sum_{i=1}^n [i,i+1,\cdots, i+k] \pmod{10^9+7}$ | 170 |
18 | alpha1022 | $$\frac{1}{n_1!n_2!}(1-y)^{n_1+n_2+2} \left(\sum_{j\ge 0} y^j(t+j)^{n_1} \right)
\left(\sum_{j\ge 0} y^j((j+1)-t)^{n_2} \right)$$ | 163 |
19 | ckiseki | | 161 |
20 | feecle6418 | gyh ak ioi | 160 |