水题,搞清楚hanoi的定义就好做了. /* 1329 */ #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #define MAXN 55 int b[MAXN]; int a[MAXN]; bool isSquare(int x) { int y = (int) sqrt(x*1.0); return y*y == x; } void init() { in…

Hanoi Tower Troubles Again! Problem Description People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with the…

找规律的题目an=an-1+(i+i%2)/2*2; ;}…

Hanoi Tower Troubles Again! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 602    Accepted Submission(s): 418 Problem Description People stopped moving discs from peg to peg after they know the…

链接:ZOJ1239 Hanoi Tower Troubles Again! Description People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with…

People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with the Hanoi Tower. Mr.S invented a little game on it.…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

DP/四边形不等式 裸题环形石子合并…… 拆环为链即可 //HDOJ 3506 #include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define…

DP/四边形不等式 这题跟石子合并有点像…… dp[i][j]为将第 i 个点开始的 j 个点合并的最小代价. 易知有 dp[i][j]=min{dp[i][j] , dp[i][k-i+1]+dp[k+1][j-(k-i+1)]+w(i,k,j)} (这个地方一开始写错了……) 即,将一棵树从k处断开成(i,k)和(k+1,i+j-1)两棵树,再加上将两棵树连起来的两条树枝的长度w(i,k,j) 其中,$ w(i,k,j)=x[k+1]-x[i]+y[k]-y[i+j-1] $ 那么根据四边形…

DP/四边形不等式 要求将一个可重集S分成M个子集,求子集的极差的平方和最小是多少…… 首先我们先将这N个数排序,容易想到每个自己都对应着这个有序数组中的一段……而不会是互相穿插着= =因为交换一下明显可以减小极差 然后……直接四边形不等式上吧……这应该不用证明了吧? MLE了一次:这次的w函数不能再开数组去存了……会爆的,直接算就行了= =反正是知道下标直接就能乘出来. 数据比较弱,我没开long long保存中间结果居然也没爆……(只保证最后结果不会爆int,没说DP过程中不会……) //H…