随笔分类 - AtCoder
摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc258_h "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc258/tasks/abc258_h "AtCoder 传送门") 不错的矩阵快速幂优
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc153_d "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc153/tasks/arc153_d "AtCoder 传送门") 又浪费一道好题 [AtCoder 传送门](https://atcoder.jp/contests/arc148/tasks/arc148_e "AtCoder 传送门") 是一道不错的计数。
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc161_e "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc161/tasks/arc161_e "AtCoder 传送门") 给构造题提供了一种
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc303_g "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc303/tasks/abc303_g "AtCoder 传送门") 经典题,考虑区间
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摘要:洛谷传送门 AtCoder 传送门 好题。 这种题一般可以考虑,观察最优解的性质,对于性质计数。 发现如果 \(n,m\) 均为偶数,可以放满。就是类似这样: #.#.#. .#.#.# #.#.#. .#.#.# 因此答案就是 \(2\)。 如果 \(n,m\) 有一个为偶数,不妨假设 \(n\)
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc146_d "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc146/tasks/arc146_d "AtCoder 传送门") 考虑直接增量构造。
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc193_f "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc193/tasks/abc193_f "AtCoder 传送门") 复习一下最小割。
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc146_c "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc146/tasks/arc146_c "AtCoder 传送门") 好可爱的题啊。 没
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc267_h "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc267/tasks/abc267_h "AtCoder 传送门") 直接暴力跑背包的复
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc139_d "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc139/tasks/arc139_d "AtCoder 传送门") 看成方案数想了 1
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc132_e "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc132/tasks/arc132_e "AtCoder 传送门") 感觉挺 educa
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc132_f "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc132/tasks/arc132_f "AtCoder 传送门") 没见过这种在新运算
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摘要:[洛谷传送门](http://https://www.luogu.com.cn/problem/AT_arc139_c "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc139/tasks/arc139_c "AtCoder 传送门") ~~
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc132_d "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc132/tasks/arc132_d "AtCoder 传送门") 提供一个 dp 思
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_agc062_b "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/agc062/tasks/agc062_b "AtCoder 传送门") 妙妙题。 像这种最
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_abc302_h "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/abc302/tasks/abc302_h "AtCoder 传送门") 考虑如果只询问一次
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摘要:这题太神仙了吧!感觉还不是很懂,但是尽力理一下思路。 设 $t_x$ 为最大的 $j$ 使得 $i_j = x$,不存在则 $t_x = 0$。 设 $1 \sim n$ 的数按照 $t$ 从小到大排序后是 $p_1, p_2, ..., p_n$,设 $q_i$ 为 $i$ 在 $p$ 中的排名,
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc130_c "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc130/tasks/arc130_c "AtCoder 传送门") 分类讨论,但是写起
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摘要:[洛谷传送门](https://www.luogu.com.cn/problem/AT_arc133_e "洛谷传送门") [AtCoder 传送门](https://atcoder.jp/contests/arc133/tasks/arc133_e "AtCoder 传送门") 其实是套路题,但是
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