Log2预处理
int lg[100001];
inline void log_2()
{
for (int i = 1; i <= n; i++)
{
lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
}
}
还可以 用这种方法
int Log2[100001];
inline void log_2()
{
for (int i = 2; i <= n; ++i)
Log2[i] = Log2[i / 2] + 1;
}
神之操作
std :: __lg(size_t)
(它是 的)
本文来自博客园,作者:蒟蒻orz,转载请注明原文链接:https://www.cnblogs.com/orzz/p/18122180