C-easymath\[\Pia_i\le2024^b\\\log_2(\Pi2^{k_i})\le\log_2(2024^b)\\\sumk_i\leb\log_22024\]因此答案就是\(b=\frac{\sumk_i}{\log_22024}\)#include<bits/stdc++.h>usingnamespacestd;usingi32=int32_t;usingi64