bi=∑j=1n(i,j)c−d∗idjdzjb_i=\sum_{j=1}^n(i,j)^{c-d}*i^dj^dz_jbi=j=1∑n(i,j)c−d∗idjdzj
biid=∑j=1nzjjd(i,j)c−d\frac{b_i}{i^d}=\sum_{j=1}^nz_jj^d(i,j)^{c-d}idbi=j=1∑nzjjd(i,j)c−d记(i,j)=x,Bi=biid,Zi=ziid,fi(d)=∑(i,j)=dZj,Fi(d)=∑d∣(i,j)Zj(i,j)=x,B_i=\frac{b_i}{i^d},Z_i=z_ii^d,f_i(d)=\sum_{(i,j)=d}Z_j,F_i(d)=\sum_{d|(i,j)}Z_j(i,j)=x,Bi=id
UOJ62 怎样跑的更快
最新推荐文章于 2025-06-14 19:22:02 发布