题目传送门 /* 线段树的单点更新:有一个交叉更新,若rank=1,or:rank=0,xor 详细解释:http://www.xuebuyuan.com/1154895.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <cmath> #include <…

题目传送门 /* 题意:对于长度为x的子序列,每个序列存放为最小值,输出长度为x的子序列的最大值 set+线段树:线段树每个结点存放长度为rt的最大值,更新:先升序排序,逐个添加到set中 查找左右相邻的位置,更新长度为r - l - 1的最大值,感觉线段树结构体封装不错! 详细解释:http://blog.csdn.net/u010660276/article/details/46045777 其实还有其他解法,先掌握这种:) */ #include <cstdio> #include &l…

给你一个序列,让你划分成K段,每段的价值是其内部权值的种类数,让你最大化所有段的价值之和. 裸dp f(i,j)=max{f(k,j-1)+w(k+1,i)}(0<=k<i) 先枚举j,然后枚举i的时候,用线段树进行优化,对a(i)上一次出现的位置到i之间的f(k,j-1)的答案进行+1,然后求个i的前缀max. 要注意线段树区间加的时候其实要包含上0. #include<cstdio> #include<algorithm> #include<cstring&g…

B. The Bakery time limit per test 2.5 seconds memory limit per test 256 megabytes input standard input output standard output Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven whic…

 The Bakery time limit per test 2.5 seconds memory limit per test 256 megabytes input standard input output standard output Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which…

D. The Bakery   Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which can bake several types of cakes, and opened the bakery. Soon the expenses started to overcome the income, so…

/* CodeForces 834C - The Meaningless Game [ 分析,数学 ] | Codeforces Round #426 (Div. 2) 题意: 一对数字 a,b 能不能被表示为 a = x^2 * y , b = x * y^2 分析: 看出题意就差不多可以直接上了 a^2 = x^4 * y^2 , b = x * y^2 x^3 = a^2/b 同理 y^3 = b^2/a 随便验证一下,开三次方这个,要么预打表,要么pow() 完微调 */ #includ…

D. The Bakery time limit per test 2.5 seconds memory limit per test 256 megabytes input standard input output standard output Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven whic…

题目链接:http://codeforces.com/contest/834/problem/D 题意:给定一个长度为n的序列和一个k,现在让你把这个序列分成刚好k段,并且k段的贡献之和最大.对于每一段的贡献为该段有多少个不同的数字. 思路:考虑dp, dp[k][i]表示前i个数切k段的答案,那么dp[k][i]=max(dp[k-1][j]+color(j+1,i)) [1<=j<i], [color(l,r)表示区间[l,r]有多少个不同的数字] ,由于第k行的dp值只会影响到k+1行的…

A:考虑每个质因子,显然要求次数之和是3的倍数,且次数之差的两倍不小于较小的次数.对于第一个要求,乘起来看开三次方是否是整数即可.第二个取gcd,两个数分别除掉gcd,之后看两个数的剩余部分是否都能被gcd整除即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> us…