新鲜热乎的题 Codeforce 1175 D. 题意:给出一个长度为$n$的序列$a$,你需要把它划分为$k$段,每一个元素都需要刚好在其中一段中.分好之后,要计算$\sum_{i=1}^{n} (a_i \centerdot f(i))$,使这个值最大.其中$f(i)$代表第$i$个元素被分在第几段中.简单来说,就是每个元素被分在第几段中就需要乘上几,使序列的和最大. 假设$sum$是$a$的前缀和,$k$个分割点分别为$l_j$,将序列分割为$[1, l_1],[l_1 + 1, l_2]…

A. Lesha and array splitting time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output One spring day on his way to university Lesha found an array A. Lesha likes to split arrays into several parts.…

题目链接 Lesha and array splitting 设s[i][j]为序列i到j的和,当s[i][j]≠0时,即可从i跳到j+1.目标为从1跳到n+1,所以按照题意暴力即可. #include <bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i(a); i <= (b); ++i) #define dec(i,a,b) for(int i(a); i >= (b); --i) + ; stru…

C. Array Splitting You are given a sorted array…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output One spring day on his way to university Lesha found an array A. Lesha likes to split arrays into several parts. This time Lesha decided to split…

You are given an array a1,a2,…,ana1,a2,…,an and an integer kk. You are asked to divide this array into kk non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray. Let f(i)f(i) be the index of subarray th…

You are given an array…

D. Beautiful Array You are given an array aa consisting of nn integers. Beauty of array is the maximum sum of some consecutive subarray of this array (this subarray may be empty). For example, the beauty of the array [10, -5, 10, -4, 1] is 15, and th…

10^4以内只由4和7构成的数字只有31种,那么做法就很简单了,求出每个数字与其最接近的幸运数的差值,然后建立线段树,线段树维护区间最小值和最小值个数,如果操作过程中最小值<0,那么就去对差值进行暴力修改,直到区间差值>=0,很明显线段树每个叶子节点不会被修改超过31次,询问操作的话差值=0的数字就是幸运数了.复杂度O(31nlogn) 代码 #include <algorithm> #include <iostream> #include <sstream>…

应该是做麻烦了,一开始还没A(幸好上一次比赛水惨了) #include<bits/stdc++.h> #define lowbit(x) x&(-x) #define LL long long #define N 200005 #define M 1000005 #define mod 2147483648LL #define inf 0x7ffffffff using namespace std; inline int ra() { ,f=; char ch=getchar(); ;…