codeforces 4A Watermelon

Watermelon Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on CodeForces. Original ID: 4A 64-bit integer IO format: %I64d      Java class name: (Any) Prev Submit Status Statistics Discuss Next Type: None None Graph Theory      2-

Codeforces印象 这两天抽时间去codeforces体验了一把. 首先,果然有众多大牛存在,很多名人一直参加每周一次的比赛,积分2000+,并参与出题. 另外,上面题目很多,估计至少一千题.比赛结束后,题目将转为练习题,可以持续尝试.每道题目都有标签,如greedy, math, matrices等等,可以点击相应的标签只做相关的题目.可惜我做了好几道之后才发现. 这次解决的题目 首次尝试,这次做的几个都是选的完成人数最多的,比较简单,但还是有些东西是从中新学习到的.以后最好分类练习.

水题一个,就是记录下已经做过了. 题目大意: 给出一个数,问是否能分成两个偶数. 下面是代码: #include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h> #include <stdlib

题目大意&链接: http://codeforces.com/problemset/problem/718/E 给一个长为n(n<=100 000)的只包含‘a’~‘h’8个字符的字符串s.两个位置i,j(i!=j)存在一条边,当且仅当|i-j|==1或s[i]==s[j].求这个无向图的直径,以及直径数量. 题解:  命题1:任意位置之间距离不会大于15. 证明:对于任意两个位置i,j之间,其所经过每种字符不会超过2个(因为相同字符会连边),所以i,j经过节点至多为16,也就意味着边数至多

题目链接:http://codeforces.com/problemset/problem/124/A Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less than a people standing in front of him and no more than b people standing b

Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or  - 1. To pass th

链接: http://codeforces.com/problemset/problem/506/A 题意: 给出30000个岛,有n个宝石分布在上面,第一步到d位置,每次走的距离与上一步的差距不大于1,问走完一路最多捡到多少块宝石. 题解: 容易想到DP,dp[i][j]表示到达 i 处,现在步长为 j 时最多收集到的财富,转移也不难,cnt[i]表示 i 处的财富. dp[i+step-1] = max(dp[i+step-1],dp[i][j]+cnt[i+step+1]) dp[i+st

You have n devices that you want to use simultaneously. The i-th device uses ai units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·ai units of power. The i-th device currently has bi units of power store

题目链接:Educational Codeforces Round 21 G. Anthem of Berland 题意: 给你两个字符串,第一个字符串包含问号,问号可以变成任意字符串. 问你第一个字符串最多包含多少个第二个字符串. 题解: 考虑dp[i][j],表示当前考虑到第一个串的第i位,已经匹配到第二个字符串的第j位. 这样的话复杂度为26*n*m*O(fail). fail可以用kmp进行预处理,将26个字母全部处理出来,这样复杂度就变成了26*n*m. 状态转移看代码(就是一个kmp