吐槽一下,这个OJ的题目真的是阅读理解题.代码非常短,就是题目难理解.心累. 传送门:点我 Chocolate bars It is hard to overestimate the role of chocolate bars in traditional programming competitions. Firstly, the nutritional content of chocolate significantly increases the number of brilliant…

P2983 [USACO10FEB]购买巧克力Chocolate Buying 题目描述 Bessie and the herd love chocolate so Farmer John is buying them some. The Bovine Chocolate Store features N (1 <= N <= 100,000) kinds of chocolate in essentially unlimited quantities. Each type i of choc…

题目链接 题目大意 有一个长度为n的全排列,你可以询问2n次,要你经过这个2n次的询问后,求出这个全排列 询问定义为:输入"? i j"输出\(p_{i} mod p_{j}\) 题目思路 比赛的时候想了很久都没思路,一直想的是用O(n)的方法确定出这个全排列n的位置,然后再用O(n)用其他位置的数去模它.结果一直没写出来. 这个题目的思路其实就是两个数互模,那么大的的那个模数必然是这两个数中的小的那个数 然后题目就变得简单起来,每2次查询就能确定一个值,然后再用另外一个不确定的值去和…

中文题意: 给你两个长度为 \(n\) 的01串 \(s,f,\)有 \(q\) 次询问. 每次询问有区间 \([\ l,r\ ]\) ,如果 \([\ l,r\ ]\) 同时包含\(0\)和\(1\),则询问终止,否则你可以将区间\([\ l,r\ ]\) 内严格小于 \(len_{lr}\) 的数字. 问是否可以使得询问不终止,且经过 \(q\) 次询问后可以将\(s\)改为\(f\). 前置知识: 线段树 没了 思路: 发现没法正序推过去(反正我不会),考虑根据询问逆推. 那么对于 \(f…

C. Alice, Bob and Chocolate 题目连接: http://codeforces.com/contest/6/problem/C Description Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one f…

A.Regular bracket sequence A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting…

题目链接 CF 633 F. The Chocolate Spree 题解 维护子数答案 子数直径 子数最远点 单子数最长直径 (最长的 最远点+一条链) 讨论转移 代码 #include<vector> #include<cstdio> #include<algorithm> #define gc getchar() #define pc putchar #define int long long inline int read() { int x = 0,f = 1…

http://codeforces.com/contest/490/problem/D 好神的一题,不会做.. 其实就是将所有的质因子找出来,满足: 最终的所有质因子的乘积相等 但是我们只能操作质因子2和3,那么我们就要将任意一边质因子多的先约掉.且先执行3,因为操作是2/3嘛.. 所以筛掉3后再筛2,然后判断面积是否等即可 #include <cstdio> #include <cstring> #include <cmath> #include <string…

Problem description Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololat…

题意:给出a1*b1和a2*b2两块巧克力,每次可以将这四个数中的随意一个数乘以1/2或者2/3,前提是要可以被2或者3整除,要求最小的次数让a1*b1=a2*b2,并求出这四个数最后的大小. 做法:非常显然仅仅跟2跟3有关.所以s1=a1*b1,s2=a2*b2,s1/=gcd(s1,s2),s2/=gcd(s1,s2),然后若s1跟s2的质因子都是2跟3,那么就有解.之后暴力乱搞就好了. #include<map> #include<string> #include<cs…