题目传送门 题意 这道题被某大佬改编拿来出成考试题,是长这个样子的: 好的,其实这才是真正的题意: 给定初始序列和最终序列,每次选择一个数变成自己和相邻2个数的和.问初始序列是否可以变为最终序列,若可以,问最少需要多少次. 分析 发现这道题有很多种操作方式,就算是写搜索暴力都不是很好写. 正难则反,考虑从末状态到初状态,就是一直减去左右两边的数. 如果中间的数大于两边的数之和,那么中间那个数一定要被操作(设$a,b,c$分别为$i-1,i,i+1$上的$B$值) 而且在$b>a+c$条件不被破坏…

Atcoder 全是神仙题-- 先变成能不能从 \(b\) 到 \(a\).操作变成一个数减掉旁边两个数. 考虑里面最大的且不和 \(a\) 中相等的那个数.它两边的数此时都不能操作,否则就减到非正数了. 而且应该要一直对这一位进行操作,直到等于 \(a_i\) 或者不是最大值为止.这样两边的数才能操作,或者真正确定无解. 用个堆模拟即可. 我的代码中复杂度--大概是两个 \(\log\) 吧.(辗转相除算一个) #include<bits/stdc++.h> using namespace…

从后往前做,每次将\(B_i\)减去相邻两个数,注意如果最大的数没有变成初始状态,那么肯定要减,否则相邻两边的就减不了,所以用堆维护.根据辗转相除的复杂度,\(O(n\log^2 n)\). #include<bits/stdc++.h> #define Rint register int #define MP make_pair #define fi first #define se second using namespace std; typedef long long LL; type…

题目大意 给你一个序列a和序列b 每次操作是a[i]+=a[i-1]+a[i+1] 问a经过最少几次操作可以得到b 分析 用堆维护a 每次取出最大的 撤销操作直到不能撤销 将新数放入堆 不断维护即可 代码 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<cctype> #include…

D. Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day Anna got the following task at school: to arrange several numbers in a circle so that any two neighboring numbers differs…

D. Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day Anna got the following task at school: to arrange several numbers in a circle so that any two neighboring numbers differs…

最前面: AT的题都很有思维难度,总结一下一些AT的常规操作 1.对于有操作的题目,如果正面推不行的话考虑倒推,将操作转化,寻找更好的性质 2.模型转化,看到某一种的计算的式子,需要考虑有没有更简化的模型可以达到相同的效果 3.补集转化,正难则反 4.分析题目性质,计数题要找到一些限制条件或者构造方案使得计数不重不漏 目标:稳定三题 比例:D:0/1  C:3/6 B:2/3 10.12 - 10.22 AGC47:Solved A,B,C,D Aim E AGC46:Solved A,B,C,…

Node.js is a platform for building fast and scalable server applications using JavaScript. Node.js is the runtime and NPM is the Package Manager for Node.js modules. VS Code has support for the JavaScript and TypeScript languages out-of-the-box as we…

  Prime Ring Problem  A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always…

Meeting Minutes Merge Wordlist & Word Recite entry. (P0) – Done. Remove "Word Challenge" Button on user's wordlist expansion. (P0) – Done. Move "1453, 0" numbers inside the circle.(P2) – Done. Add "我认识" or "我不认识"…

Jump Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1452 Integers 1, 2, 3,..., n are placed on a circle in the increasing order as in the following figure. We want to construct a sequence from these…

Description   A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1. I…

Integers 1, 2, 3,..., n are placed on a circle in the increasing order as in the following figure. We want to construct a sequence from these numbers on a circle. Starting with the number 1, we continually go round by picking out each k-th number and…

Description   A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1.…

Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1. Inp…

UVA - 524 Prime Ring Problem Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of number…

素数环(暴力)(紫书194页) Description   A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle shoul…

Game Description: 1. In a closed door, there will be a circle, and 30 numbers in the circle. 2. Each team (about 10 people) need to take number out of the circle by ascend order. 3. Each team has 30 minutes and 3 tries, each try is limited in 90 seco…

题意  把1到n这n个数以1为首位围成一圈  输出全部满足随意相邻两数之和均为素数的全部排列 直接枚举排列看是否符合肯定会超时的  n最大为16  利用回溯法 边生成边推断  就要快非常多了 #include<cstdio> using namespace std; const int N = 50; int p[N], vis[N], a[N], n; int isPrime(int k) { for(int i = 2; i * i <= k; ++i) if(k % i == 0)…

本文转自:https://code.visualstudio.com/docs/nodejs/nodejs-tutorial Node.js tutorial in Visual Studio Code Node.js is a platform for building fast and scalable server applications using JavaScript. Node.js is the runtime and NPM is the Package Manager for…

Contest2073 - 湖南多校对抗赛(2015.04.06) Problem A: (More) Multiplication Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 85  Solved: 47[Submit][Status][Web Board] Description Educators are always coming up with new ways to teach math to students. In 2011, a…

Preface 这篇咕了可能快一个月了吧,正好今天晚上不想做题就来补博客 现在还不去复习初赛我感觉我还是挺刚的(微笑) A - Dividing a String 考虑最好情况把每个字符串当作一个来看,考虑不合法的情况怎么处理 可以很容易地发现再怎么差我长度分成\(1,2,1,2,\cdots\)的样子就好了,因此字符串最长为\(2\) 贪心地把后面的数和前面合并即可 #include<cstdio> #include<cstring> #define RI register in…

从这里开始 题目目录 Problem A Dividing a String 猜想每段长度不超过2.然后dp即可. 考虑最后一个长度大于等于3的一段,如果划成$1 + 2$会和后面相同,那么划成$2 + 1$,如果前一段和前面相同,那么把前一段和前面合并.每次操作后段数都不会减少.所以存在一种最优方案使得每段长度不超过2. Code #include <bits/stdc++.h> using namespace std; typedef bool boolean; const int N =…

A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example,the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionallynumbers can of course be ordered in size. The first few…

Alexandra and Prime Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1658    Accepted Submission(s): 565 Problem Description Alexandra has a little brother. He is new to programming. One…

我这种maintain写法好zz.考试时获得了40pts的RE好成绩 In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2). DZY loves Fibonacci numbers very much. Today DZY gives you an array con…

2019 杭电多校 8 1003 题目链接:HDU 6659 比赛链接:2019 Multi-University Training Contest 8 Problem Description Acesrc is a famous mathematician at Nanjing University second to none. Playing with interesting numbers is his favorite. Today, he finds a manuscript whe…

Codeforces Round #597 (Div. 2 Consider the set of all nonnegative integers: 0,1,2,-. Given two integers a and b (1≤a,b≤104). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on. Each number is painted whi…

题意:博弈题面  给出一个数字序列 (>=11)  有两个人任意删除数字 直到 数字只剩下11位 如果删除后的数字串开头是8那么就是第一个赢 否则就是第二个人赢 第一个人先手  数字序列一定是奇数长度 思路: 首先计算一共走多少步 第二个人想赢只有以下两种方法 想法1:如果第二个人能把8都删掉 那么第二个人肯定赢 想法2: 如果删不掉 那么第二个人肯定从前到后尽可能得删掉8 这样使得第一个人的步数不足删除从前到后 步数+1那个8 的其他字符 否则就是第一人赢,直接模拟即可 (比赛的时候一直WA…

妙妙技巧题 题目描述 题目大意 一个圆环上有n只蚂蚁,它们会按照顺时针或者逆时针行走.如果有蚂蚁相遇它们就会掉头(不一定在整数时间掉转).问最后每只蚂蚁的位置. 题目分析 以前在luogu上做过一道类似的掉头问题,于是考试时候就很大意地把它压在最后做了…… 我们考虑一下,这n只蚂蚁的相对位置其实是不会改变的,并且我们可以算出最终所有蚂蚁的位置. 那应该如何计算第一只蚂蚁最后的顺序呢? 想象在n-1到0之间有一只哨兵蚂蚁,它是不会移动的.当有蚂蚁从左边或右边穿过它的时候,我们的一号蚂蚁相当于次序+…