题意:问随机生成一个长度为m(m<=1000)长度的字符串,出现某个子串s的概率是多少. /* KMP+DP 设f[i][j]表示A生成到第i位,此时B串匹配到第j位的概率. 转移方程为f[i+1][k]+=f[i][j]*lv(枚举的第i位生成字符的概率) k位如果第i位生成该字符,那么它能和B串匹配到的地方. */ #include<cstdio> #include<iostream> #include<cstring> #define N 1010 usin…

J - Infinite monkey theorem Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3689 Appoint description:  System Crawler  (2014-11-09) Description Could you imaging a monkey writing computer progra…

Infinite monkey theorem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)http://acm.hdu.edu.cn/showproblem.php?pid=3689 Problem Description Could you imaging a monkey writing computer programs? Surely monkeys are sma…

Infinite monkey theorem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1702 Accepted Submission(s): 882 Problem Description Could you imaging a monkey writing computer programs? Surely monkeys are…

Description Could you imaging a monkey writing computer programs? Surely monkeys are smart among animals. But their limited intelligence is no match for our human beings. However, there is a theorem about monkeys, and it states that monkeys can write…

Infinite monkey theorem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1703    Accepted Submission(s): 883 Problem Description Could you imaging a monkey writing computer programs? Surely monke…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3689 黄老师说得对,题目只有做wa了才会有收获,才会有提高. 题意:一个猴子敲键盘,键盘上有n个键,猴子敲第i个键的概率是p[i],问敲m次后形成的字符串里出现给定串的概率是多少. 这实际上就跟那个ac自动机转为trie图然后dp一样的. 类似的题目有POJ 2778,这篇题解不错:http://blog.csdn.net/luyuncheng/article/details/8643001 只不过…

题链: http://acm.hdu.edu.cn/showproblem.php?pid=3689题解: KMP,概率dp (字符串都从1位置开始) 首先对模式串S建立next数组. 定义dp[i][j]表示猴子打的串长度为i,且该串的后缀与模式串最多匹配到j位置的概率. 显然dp[0][0]=1, 考虑如何转移: 枚举下一个打出的字符为c,然后用kmp的next数组找到模式串中可以继续匹配的位置k. 即:k=j+1; while(k&&S[k]!=c) k=next[k]; 然后将dp…

意甲冠军: 给n快报,和m频率. 然后进入n字母出现的概率 然后给目标字符串str 然后问m概率倍的目标字符串是敲数量. 思维: AC自己主动机+可能性dp简单的问题. 首先建立trie图,然后就是状态转移了 dp版本号: dp三重循环变量次数,节点数,和字母数 代码: #include"cstdlib" #include"cstdio" #include"cstring" #include"cmath" #include&q…

这题由于是一个单词,其实直接kmp+dp也无妨.建立自动机当然也是可以的.设dp[i][j]表示匹配到第i个字母的时候,在单词中处于第j个位置的概率,因此最终的答案是dp[0~m][len],m是输入的长度,len是单词的长度.转移方程见代码,即在一个节点的位置时,枚举下一步的走法,乘以这种走法的概率即是对下一个状态的概率贡献. 其实自动机的fail指针跳转的位置即是kmp中nxt数组跳转的位置. 代码如下: { ] != word[i]) j = nxt[j]; ] == word[i]) j…