题意:有一个长度为\(n\)的序列,你每次可以对序列重新排序,然后花费\(1\)使某个元素加减\(1\),多次操作后使得新序列满足\(a_{i}=c^i\),\(c\)是某个正整数,求最小花费. 题解:先排序,我们可以直接枚举\(c\),然后模拟维护一个最小值就好了. 代码: int n; int a[N]; int main() { //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); n=read(); for(int i=1;i<=n;…

题目链接:Power Sequence 题意: 给你n个数vi,你可以对这个序列进行两种操作 1.可以改变其中任意个vi的位置,无成本 2.可以对vi进行加1或减1,每次操作成本为1 如果操作之后的vi(设v数组下标从1到n)满足:如果存在一个数c,使得每一个vi都满足vi==ci 你需要输出这个满足题意得vi数组构成所需的最小成本 题解: 题目要求1<=ai<1e9,那么可以说所有vi都加起来得数量级是1e9,那么也就说如果cn,那么c这个数就不可取 因为我们c可以取1,那么成本最大也是在1…

比赛链接:https://codeforces.com/contest/1397 A. Juggling Letters 题意 给出 $n$ 个字符串,可在字符串间任意移动字母,问最终能否使这 $n$ 个字符串相同. 题解 如果可以,因为 $n$ 个字符串相同,所以每个字母的数量一定是 $n$ 的倍数. 代码 #include <bits/stdc++.h> using namespace std; void solve() { int n; cin >> n; vector<…

B. Lipshitz Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/601/problem/B Description A function  is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| hold…

A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/675/problem/A Description Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a), and the difference between…

链接: https://codeforces.com/contest/1265/problem/D 题意: An integer sequence is called beautiful if the difference between any two consecutive numbers is equal to 1. More formally, a sequence s1,s2,-,sn is beautiful if |si−si+1|=1 for all 1≤i≤n−1. Trans…

Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a), and the difference between any two neighbouring elements is equal to c (si - si - 1 = c). In particular, Vasya won…

. Hungry Sequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money,…

题意:n个点, 坐标已知,其中横坐标为为1~n. 求区间[l, r] 的所有子区间内斜率最大值的和. 首先要知道,[l, r]区间内最大的斜率必然是相邻的两个点构成的. 然后问题就变成了求区间[l, r]内所有子区间最大值的和. 这个问题可以利用单调栈来做. 每次找到当前点左面第一个大于当前值的点, 然后更新答案. 姿势很多. import java.io.BufferedInputStream; import java.io.BufferedOutputStream; import java.…

题意: 给一个数组,给你一个k,找出两个数字的积可以变成xk的数对对数 解析: 当且仅当,两个数进行质因子分解后每个因子的个数都是k的倍数个就说明这是满足条件的一对,可以让每个因子个数%k用map找对应的数. 代码: #include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <vector> #include <cstdio…