题意:给一个初始值1,每步操作将1替换为01,将0替换为10.问N步操作后有多少对连续的0. 解法:f[i]表示第i步后的答案.可以直接打表发现规律--奇数步后,f[i]=f[i-1]*2-1;偶数步后,f[i]=f[i-1]*2+1;至于原因--我只能简单说一点.第i步后的答案可由i-1步后的"01"+"1"+"0"的个数推出,而"01"*2+"1"+"0"=01串的总个数.用x表示i…

题意:定义一棵树的所有非叶节点都恰好有n个儿子为严格n元树.问深度为d的严格n元树数目. 解法:f[i]表示深度为<=i的严格n元树数目.f[i]-f[i-1]表示深度为i的严格n元树数目.f[i]=f[i-1]^n+1.d层的严格n元树可分解为1个根节点和n棵d-1层的严格n元树.利用乘法原理,再加上子树为空的一种情况. P.S.同样要注意递推的思想! 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring&…

Computer Transformation Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4561 Accepted: 1738 Description A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simul…

Computer Transformation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6946    Accepted Submission(s): 2515 Problem Description A sequence consisting of one digit, the number 1 is initially wri…

Computer Transformation http://acm.hdu.edu.cn/showproblem.php?pid=1041 Problem Description A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each…

Computer Transformation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8367    Accepted Submission(s): 3139 Problem Description A sequence consisting of one digit, the number 1 is initially wri…

Computer Transformation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8688    Accepted Submission(s): 3282 Problem Description A sequence consisting of one digit, the number 1 is initially wr…

POJ.3624 Charm Bracelet(DP 01背包) 题意分析 裸01背包 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 13000 #define nnmax 3500 using namespace std; int dp[nmax]; int w[nnmax],d[nnmax]; int main…

H - Computer Transformation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice _ Appoint description:  System Crawler  (Oct 10, 2016 1:02:59 PM) Description A sequence consisting of one digit, the numb…

POJ 2995 Brackets 区间DP 题意 大意:给你一个字符串,询问这个字符串满足要求的有多少,()和[]都是一个匹配.需要注意的是这里的匹配规则. 解题思路 区间DP,开始自己没想到是区间DP,以为就是用栈进行模拟呢,可是发现就是不大对,后来想到是不是使用DP,但是开始的时候自己没有推出递推关系,后来实在想不出来看的题解,才知道是区间DP,仔细一想确实是啊. 下面就是状态转移方程: \[ \begin{cases}dp[i][j] &=& dp[i+1][j-1]+if(str…