POJ-1644 To Bet or Not To Bet(概率DP)】的更多相关文章

poj double 就得交c++,我交G++错了一次 题目:http://poj.org/problem?id=2151 题意:ACM比赛中,共M道题,T个队,pij表示第i队解出第j题的概率 问 每队至少解出一题且冠军队至少解出N道题的概率. 每队均至少做一题的概率P1 减去 每队做题数均在1到N-1之间的概率P2 设dp[i][j][k]表示第i个队在前j道题中解出k道的概率 则: dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p…
题目链接:http://poj.org/problem?id=2151 题意: 一次ACM比赛,有t支队伍,比赛共m道题. 第i支队伍做出第j道题的概率为p[i][j]. 问你所有队伍都至少做出一道,并且有队伍做出至少n道的概率. 题解: 关于[至少]问题的表示. 对于每一支队伍: mst[i][j] = P(第i支队伍做出至多j道题) 则 P(第i支队伍做出至少j道题) = 1 - mst[i][j-1] 对于所有队伍: P(所有队伍至少答出一题) = ∏ (1 - mst[i][0]) P(…
题意:给出m.t.n,接着给出t行m列,表示第i个队伍解决第j题的概率. 现在让你求:每个队伍都至少解出1题,且解出题目最多的队伍至少要解出n道题的概率是多少? 思路:求补集. 即所有队伍都解出题目的概率,减去所有队伍解出的题数在1~n-1之间的概率 这里关键是如何求出某个队伍解出的题数在1~n-1之间的概率,采用dp的方法: 用p(i,j)表示前i道题能解出j道的概率,有p(i,j)=p(i-1,j)*(1-p(i))+p(i-1,j-1)*p(i)p(i)表示解出第i题的概率. #inclu…
To Bet or Not To Bet Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 1668 Accepted: 541 Description Alexander Charles McMillan loves to gamble, and during his last trip to the casino he ran across a new game. It is played on a linear seque…
UVA 1541 - To Bet or Not To Bet 题目链接 题意:这题题意真是神了- -.看半天,大概是玩一个游戏,開始在位置0.终点在位置m + 1,每次扔一个硬币,正面走一步,反面走两步,走到的步上有4种情况: 1.向前走n步 2.向后走n步 3.停止一回合 4.无影响 问能在t次机会内,走到终点m + 1(假设跃过也算走到了)的概率.大于0.5.等于0.5,小于0.5相应不同输出 思路:题意懂了就好办了.事实上就是递推就能够了dp[i][j]表示第i次机会,落在j步的概率.然…
Alexander Charles McMillan loves to gamble, and during his last trip to the casino he ran across a new game. It is played on a linear sequence of squares as shown below. A chip is initially placed on the Start square. The player then tries to move th…
id=3071">http://poj.org/problem? id=3071 大致题意:有2^n个足球队分成n组打比赛.给出一个矩阵a[][],a[i][j]表示i队赢得j队的概率.n次比赛的流程像这样action=showproblem&problemid=2304">France \'98. 问最后哪个队最可能得冠军. 思路:概率dp问题.ans[i][j]表示第i轮中j队获胜的概率. #include <stdio.h> #include &l…
Check the difficulty of problems Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8903   Accepted: 3772 Description Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually exp…
[POJ 2750] Potted Flower(线段树套dp) Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4566   Accepted: 1739 Description The little cat takes over the management of a new park. There is a large circular statue in the center of the park, surrou…
Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug…