题目传送门 /* 题意:任意排列第一个字符串,使得有最多的不覆盖a/b字符串出现 字符串处理/贪心:暴力找到最大能不覆盖的a字符串,然后在b字符串中动态得出最优解 恶心死我了,我最初想输出最多的a,再最多的b,然而并不能保证是最多的:( */ #include <cstdio> #include <cstring> #include <string> #include <iostream> #include <algorithm> #includ…

题目传送门 /* 水题:开个结构体,rk记录排名,相同的值有相同的排名 */ #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <iostream> #include <queue> #include <map> #include <vector>…

E. GukiZ and GukiZiana Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/E Description Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given a…

C. GukiZ hates Boxes Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/C Description Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way. In total there…

B. ZgukistringZ Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/B Description Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one. GukiZ has strings a,…

A. GukiZ and Contest Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/A Description Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has de…

题意与分析(CodeForces 551B) 这他妈哪里是日常训练,这是日常弟中弟. 题意是这样的,给出一个字符串A,再给出两个字符串B,C,求A中任意量字符交换后(不限制次数)能够得到的使B,C作为子串不重叠且出现次数最多的串. 看起来很简单对吧,做法也很简单,先排序字符,然后枚举合法的B串能在A串中出现的次数,看能有几个C串,然后求个最优解就行了.是不是很简单?然后WA了十几发,各种捉bug,太杀妈了. 这份代码看起来贼简单,这是我写的第三版了.... 这种字符串题目太苦手了QAQ 代码 #…

题目地址:http://codeforces.com/contest/551/problem/D 分析下公式能够知道,相当于每一位上放0或者1使得最后成为0或者1.假设最后是0的话,那么全部相邻位一定不能全是1,由于假设有一对相邻位全为1,那么这两个的AND值为1.又由于OR值是仅仅要有1.结果就为1.所以这位结果肯定为1.所以就推出了一个dp转移方程.dp[i][j]表示第i位上的数为j时的总个数.那么有: dp[i][0]=dp[i-1][0]+dp[i-1][1]; dp[i][1]=dp…

题目地址:http://codeforces.com/contest/551/problem/E 将n平均分成sqrt(n)块,对每一块从小到大排序,并设置一个总体偏移量. 改动操作:l~r区间内,对两端的块进行暴力处理,对中间的总体的块用总体偏移量标记添加了多少.时间复杂度: O(2*sqrt(n)+n/sqrt(n)). 查询操作:对每一块二分.查找y-总体偏移量.找到最左边的和最右边的.时间复杂度:O(sqrt(n)*log(sqrt(n))). 代码例如以下: #include <ios…

D. GukiZ and Binary Operations time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We all know that GukiZ often plays with arrays. Now he is thinking about this problem: how many arrays a, of l…