codeforces 804A Find Amir   http://codeforces.com/problemset/problem/804/A /* 题意:给定n个学校,需要遍历所有学校,可从任意一点开始. i到j有(i+j) mod (n+1) 的花费,求最小花费. 考虑贪心:先走和为n+1的两处,走完后最小只能再走到和为n+1的学校,再走到和为n的 例如: 1 2 3 4 5 6 7 8 9 10 先走1->10 花费为0 再走10->2 花费为1 再走2->9 花费为0 ..…

http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the shi…

题目链接:http://codeforces.com/contest/724/problem/D 题意:给定一个字符串和一个数字m,选取一个一个子序列s,使得对于字符串中任意长度为m的子序列都至少含有s的位置(不是字符),求所有s在sort后字典序最小的那个字符串. 思路:对字符排序后,从最后一个开始贪心,判断删除该字符后是否符和题意,当删除后不符合题意时,贪心到该相同字符对应的第一个位置为止. 比如对于test3来说 排序后为a a a b b b b c c c c 删除到(b,6)时发现不…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

1.codeforces 349B    Color the Fence 2.链接:http://codeforces.com/problemset/problem/349/B 3.总结: 刷栅栏.1-9每个字母分别要ai升油漆,问最多可画多大的数字. 贪心,也有点考思维. #include<bits/stdc++.h> using namespace std; #define LL long long #define INF 0x3f3f3f3f int main() { ]; while(…

B. Urbanization 题目链接 http://codeforces.com/contest/735/problem/B 题面 Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are n people who plan to move…

地址:http://codeforces.com/problemset/problem/712/C 题目: C. Memory and De-Evolution time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Memory is now interested in the de-evolution of objects, sp…

题目传送门 /* 贪心:首先要注意,y是中位数的要求:先把其他的都设置为1,那么最多有(n-1)/2个比y小的,cnt记录比y小的个数 num1是输出的1的个数,numy是除此之外的数都为y,此时的numy是最少需要的,这样才可能中位数大于等于y */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; ; con…

题目链接: http://codeforces.com/problemset/problem/268/E E. Playlist time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 Manao's friends often send him new songs. He never listens to them right away. Instead, he compiles them into a play…