Alice and Bob are playing a simple game. They line up a row of n identical coins, all with the heads facing down onto the table and the tails upward. For exactly mm times they select any k of the coins and toss them into the air, replacing each of th…

题目链接:Coins Description Alice and Bob are playing a simple game. They line up a row of nn identical coins, all with the heads facing down onto the table and the tails upward. For exactly mm times they select any kk of the coins and toss them into the…

题目链接:https://www.jisuanke.com/contest/2870?view=challenges 题目大意:给出n个都正面朝下的硬币,操作m次,每次都选取k枚硬币抛到空中,求操作m次后,硬币向上的期望值. 思路: 1.期望跟概率还是有点不同的,期望要枚举出抛的所有的情况,然后求sigma(i * dp[][]) 2.dp[i][j]表示进行i次操作后,有j枚硬币向上的概率.这样就可以求最后的硬币向上的期望了. 3.值得注意的是,预处理的组合数要开 double 型. 代码:…

Alice and Bob are playing a simple game. They line up a row of nn identical coins, all with the heads facing down onto the table and the tails upward. For exactly mm times they select any kk of the coins and toss them into the air, replacing each of…

I. Older Brother Your older brother is an amateur mathematician with lots of experience. However, his memory is very bad. He recently got interested in linear algebra over finite fields, but he does not remember exactly which finite fields exist. For…

L. Sticky Situation While on summer camp, you are playing a game of hide-and-seek in the forest. You need to designate a “safe zone”, where, if the players manage to sneak there without being detected,they beat the seeker. It is therefore of utmost i…

All as we know, a mountain is a large landform that stretches above the surrounding land in a limited area. If we as the tourists take a picture of a distant mountain and print it out, the image on the surface of paper will be in the shape of a parti…

题目 Zeit und Raum trennen dich und mich. 时空将你我分开.B 君在玩一个游戏,这个游戏由 n 个灯和 n 个开关组成,给定这 n 个灯的初始状态,下标为 从 1 到 n 的正整数.每个灯有两个状态亮和灭,我们用 1 来表示这个灯是亮的,用 0 表示这个灯是灭的,游戏 的目标是使所有灯都灭掉.但是当操作第 i 个开关时,所有编号为 i 的约数(包括 1 和 i)的灯的状态都会被 改变,即从亮变成灭,或者是从灭变成亮.B 君发现这个游戏很难,于是想到了这样的一个…

传送门 嗯……概率期望这东西太神了…… 先考虑一下最佳方案,肯定是从大到小亮的就灭(这个仔细想一想应该就能发现) 那么直接一遍枚举就能$O(nlogn)$把这个东西给搞出来 然后考虑期望dp,设$f[i]$表示从$i$个正确选项中选择一个正确的变为$i-1$个的期望次数 那么$$f[i]=\frac{i}{n}+(1-\frac{i}{n})*(1+f[i+1]+f[i])$$ 其中$\frac{i}{n}$表示一次就选了正确的选项,$(1-\frac{i}{n})$表示按错了,那么会增加一个正…

挺不错的概率DP,看似基础,实则很考验扎实的功底 这题很明显是个DP,为什么???找规律或者算组合数这种概率,N不可能给的这么友善... 因为DP一般都要在支持N^2操作嘛. 稍微理解一下,这DP[i][j]还是不好想啊,首先是写DP[I][j]的含义 首先我们想这道题是要求一个最优决策下的期望,那么这个我们的最优决策是什么??? 决策就是:我们假设我这一次需要翻转K个硬币,我们不愿翻那些已经在正面的,而去翻那些没有在正面的 而如果剩余的反面的不足,我再去翻转正面的 那么给dp[i][j]一个含…