Approximate
CF1804F Approximate Diameter 题解
题目链接 点击打开链接 题目解法 很有意思的题，但不难 首先一个显然的结论是：算着边的加入，直径长度递减 第一眼看到误差范围是 2 倍，可以想到二分 可以观察到如果取答案为 \(\frac{n}{2}\) 可以覆盖到 \(\frac{n}{4}\)（上下取整不重要），那这样每次可以把值域范围缩小 4 ......
DBMS_STATS ORA-20011 Approximate NDV failed ORA-29913 error in executing ODCIEXTTABLEOPEN callout
DBMS_STATS ORA-20011 Approximate NDV failed ORA-29913 error in executing ODCIEXTTABLEOPEN callout 目录DBMS_STATS ORA-20011 Approximate NDV failed ORA-29 ......
Atcoder ABC307_G-Approximate Equalization 序列dp
# [AT_ABC307_G-Approximate Equalization](https://atcoder.jp/contests/abc307/tasks/abc307_g "ABC307_G") [没想到还有Approximate Equalization II ！！：AT_ABC313_ ......
Atcoder ABC313_C-Approximate Equalization 2
# [AT_ABC313_C-Approximate Equalization 2](https://atcoder.jp/contests/abc313/tasks/abc313_c "ABC313_C") ## Description: - 给定一个整数序列 $A=(A_1,A_2,···,A_ ......
[ABC307G] Approximate Equalizatio
# [ABC307G] Approximate Equalizatio ## 题意 给定一个数组 $a_i$ ，有两种操作。 1. $a_i+1,a_{i+1}-1$ 2. $a_i-1,a_{i+1}+1$ 问最少花费多少次操作能使数组的极差不超过 $1$。 ## 题解 容易发现总和不变，由于极差 ......
AtCoder Beginner Contest 307 G Approximate Equalization
[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc307_g "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc307/tasks/abc307_g "AtCoder 传送门") 考虑我们如果确定了 ......