http://acm.hust.edu.cn/vjudge/contest/view.action?cid=118254#problem/C 其实这道题目不难...就是没有仔细分析... 我们可以发现,能配对的数目是最小的那个数值-1.然后就没然后了= =…

Description Bob has traveled to byteland, he find the N cities in byteland formed a tree structure, a tree structure is very special structure, there is exactly one path connecting each pair of nodes, and a tree with N nodes has N - 1 edges. As a tra…

2016 Sichuan Province Programming Contest 代码 2016 Sichuan Province Programming Contest A. Nearest Neighbor Search \(dx\)根据\(x_0\)与\([x_1, x_2]\)位置考虑. \(dx.dy.dz\)单独考虑. B. Odd Discount 做法一:对于两个不同优惠\((i,j)\),所有方案中与两个优惠商品交集为奇数,即同时取到两种优惠的个数为\(2^{n-2}\)种.…

思路:由于只有1-6这几个数,而这几个数的最小公倍数是60,所以只需要求出60以内有多少满足条件的数即可. 再就是求出对于给定的n,求出60的倍数.然后就是怎样求的问题了. 首先可以写成如下形式:                10n-40=60*n (把最后的40个数去掉,最后在求结果的时候再加上)                n=(10n-40)/60. (结果一定可以整除) 代码如下: #include<cstdio> #define ll long long #define mod…

Counting Pair Time Limit: 1000 ms Memory Limit: 65535 kB Solved: 112 Tried: 1209 Submit Status Best Solution Back Description   Bob hosts a party and invites N boys and M girls. He gives every boy here a unique number Ni(1 <= Ni <= N). And for the g…

Problem G Good Teacher I want to be a good teacher, so at least I need to remember all the student names. However, there are too many students, so I failed. It is a shame, so I don't want my students to know this. Whenever I need to call someone, I c…

我是铁牌选手 这次比赛非常得爆炸,可以说体验极差,是这辈子自己最脑残的事情之一. 天时,地利,人和一样没有,而且自己早早地就想好了甩锅的套路. 按理说不开K就不会这么惨了啊,而且自己也是毒,不知道段错误也可以是MLE,而且我的内存就是卡了那么一点点,在比赛紧张的状态下人也变傻了吧. 这次的题目难度是两边到中间一次递增的,但是我也不懂榜怎么会那样  A Peak Time Limit: 1 Second      Memory Limit: 65536 KB A sequence of  inte…

点击打开链接 Happy Reversal Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer IO format: %lld      Java class name: Main pid=34987" rel="nofollow" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" styl…

2017-2018 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2017) 全靠 wxh的博客 补完这套.wxhtxdy! [A - Concerts] $f[i][k]$ 表示在第 $i$ 个位置刚好匹配了 $k$ 个字符.转移方程 $$ f[i][k] = \sum_{i - j > h[s[k - 1]]} f[j][k - 1] $$ 前缀和优化加滚动就行了. 好像可以直接用 $f[i][k]$ 表…

2018-2019 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2018)-E. Explosion Exploit-概率+状压dp [Problem Description] 我方有\(n\)个人,对方有\(m\)个人,每个人都有一个健康值\(h_i\),有\(d\)次攻击,每次随机从所有人中选\(1\)个人,减少其\(1\)健康值.问将对方所有人都消灭的概率是多少? [Solution] 方法\(1\): ​ 将所有人的健康值作为…