树状数组,主要是抓住要求连续1的个数.这样,初始时,相同的加1,不同的加0. 查询时,用二分搜索右边界.就是比较当前mid-l+1的值与他们之间1的个数(这可以通过树状数组求区间和得出),记录右边界即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string.h> #define lowbit(x) ((x)&a…

Problem Description In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems.  On Mars, there is programming contest, too. Each team c…

意甲冠军:N个人M通过主打歌有自己的期望,每个问题发送人玩.它不能超过随机播放的次数1,追求最大业绩预期 (1 ≤ N ≤ 10,1 ≤ M ≤ 1000). 主题链接:pid=5045" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=5045 -->>设dp[i][j]表示已出战了前i - 1道题目,已出战的人的状态序列为j,如今要做第i道题目的最大期望.则最后要求的结果为max{dp[…

Contest Problem Description In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems. On Mars, there is programming contest, too. Each…

Query Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2114    Accepted Submission(s): 735 Problem Description You are given two strings s1[0..l1], s2[0..l2] and Q - number of queries. Your ta…

Coding Contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2653    Accepted Submission(s): 579 Problem Description A coding contest will be held in this university, in a huge playground. The…

题解:我们使用一个二位数组dp[i][j]记录进行到第i个任务时,人组合为j时的最大和(这里的j我们用二进制的每位相应一个人). 详细见代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; typedef long long ll; double s[11][1010]; double dp[1010][1050];…

pid=5045">主题链接~~> 做题感悟:比赛时这题后来才写的,有点小尴尬.两个人商议着写写了非常久才写出来,I want to Powerful ,I believe me . 解题思路: 遗憾 ! 感觉领悟能力太低 .由于随意时刻随意两个人做的题不超过 1 题 ,so~>  必须是一轮一轮的来(n 道题一轮),每一个人在一轮中仅仅能做一题.假设多做一题就,有可能某个人没选,某个人选择了两道,这样就不符合规定了 . 用 1  代表选择.0  代表没选择.进行某一行的时候就…

题意:给定A,B长度相同的字符串,Q次操作,修改操作位单个字符修改,查询操作为询问从某点开始有多少连续相同的字符. 思路:我们把不相同的设为1,相同的设为0,那么询问就是找下一个为1的为位置,可以用线段树解决,可以用set的lower_bound解决,这里用bitset的Find_next函数,效率还可以. 关键是代码短,好写. 复杂度:相当于分块,块的大小为32,所以复杂度为O(L/32),L取决于下一个1的位置: 可以看讨论:http://codeforces.com/blog/entry/…

利用降幂公式..呃,还是自己去搜题解吧.知道降幂公式后,就不难了. #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define LL unsigned long long using namespace std; bool mod[100005]; LL PHI(LL P){ LL ret=1; for(LL i=2;i*i<=P;i++)…