Nothing... But... Hey! This is a OIer from Canton/China.
Currently, I'm a jonior high school student & OIer in GTYZ. There're some interesting project, some of theme are in development. Hope you can join and work with me.
Nothing... But... Hey! This is a OIer from Canton/China.
Currently, I'm a jonior high school student & OIer in GTYZ. There're some interesting project, some of theme are in development. Hope you can join and work with me.
数学题!!!
为了方便化简式子,我们首先简单的将
[
d(x, y)=y-x
]
替代
[
d(x, y)=\left{
\begin{array}{l}
0\ \ (|x - y|\le1)\
y - x\ \ (|x - y| > 1)
\end{array}
\right.
]
那么就可以令 (n = 4) 将原式(
[
\sum^{n
这道题从题面就可以看出是经典的最长不下降子序列 对于这道一点都不简单的最长连续子序列和的模板题,我们还是直接暴力,代码: 123456789101112131415161718192021#include using namespace std;int ans, n, a[1009];int main() { cin >> n; f
http://localhost:4000/post/Oiclass%20P3314%20%E9%A2%98%E8%A7%A3.html
Day -INF 知道可以来集训,感觉特别开心,因为可以合法逃学两周! Day -3 (\sim) Day 0 大年初五就从潮州回到广州,收拾了一下东西,年初七大家就到了学校。 很好,又和高中生一起住,不过这次我们来得很早,所以高二还没有开学。一开始宿舍里面只有我、includeCPP、jr_inys 三个人。所以 includeCPP
这道题又又又又又是动规(我最近在学) 传统的五步: 1. 划分阶段 这里划分的阶段是在某个格子上时的最优解 2. 确定状态和状态变量 这里的状态数组 (f) 来进行存储,用数组 (a) 储存输入的数组。状态从他前面的格子获得,这些格子可能是: [ a_{S},\ a_{S+1},\ a_{S+2}...a_{T-2},\ a
https://codingcow.netlify.app/post/oiclass%20p1479%20%E9%A2%98%E8%A7%A3
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