KYOCERA Programming Contest 2022 (AtCoder Beginner Contest 271) By xiaruize D - Flip and Adjust 很明显,贪心容易有反例 ⇒\rArr⇒ 考虑 dpdpdp dp[i][j][0/1]dp[i][j][0/1]dp[i][j][0/1] 表示当前考虑到第i位,和为j,正/反面朝上是否可行 可以O(NS)O(NS)O(NS)的完成转移 如果可行,根据dp状态就可以找到序列 Code 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384/* Name: Author: xiaruize Date:*/#include <bits/stdc++.h& ...