题目链接: http://codeforces.com/problemset/problem/507/E E. Breaking Good time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that…

传送门 E. Breaking Good time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is real…

E. Breaking Good time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is really d…

题目传送门 /* 贪心水题 */ #include <cstdio> #include <algorithm> #include <iostream> #include <cmath> #include <cstring> #include <vector> #include <set> #include <map> #include <string> using namespace std; ;…

C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that loo…

A. Amr and Music (贪心) 水题,没能秒切,略尴尬. #include <cstdio> #include <algorithm> using namespace std; +; int a[maxn], r[maxn], ans[maxn]; int cmp(int i, int j) { return a[i] < a[j]; } int main() { //freopen("in.txt", "r", stdin…

C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that loo…

B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem. Amr has a circle of radius r and center in point (x, …

A. Amr and Music time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an…

传送门 D. The Maths Lecture time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher…

B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem. Amr has a circle of radius r and center in point (x, …

题目链接: http://codeforces.com/problemset/problem/208/C C. Police Station time limit per test:2 secondsmemory limit per test:256 megabytes 问题描述 The Berland road network consists of n cities and of m bidirectional roads. The cities are numbered from 1 to…

题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一个满足条件的四个点. 题解: 首先预处理出任意两点的最短距离,用队列优化的spfa跑:O(n*n*logn) 现依次访问四个点:v1,v2,v3,v4 我们可以枚举v2,v3,然后求出v2的最远点v1,v3的最远点v4,为了保证这四个点的不同,直接用最远点会错,v1,v4相同时还要考虑次最远点来替换…

题目链接:http://codeforces.com/contest/667/problem/D 给你一个有向图,dis[i][j]表示i到j的最短路,让你求dis[u][i] + dis[i][j] + dis[j][v]的最大值,其中u i j v互不相同. 先用优先队列的dijkstra预处理出i到j的最短距离(n^2 logn).(spfa也可以做) 然后枚举4个点的中间两个点i j,然后枚举与i相连节点的最短路dis[u][i](只要枚举最长的3个就行了),接着枚举与j相连节点的最短路…

题目链接: 题目 D. Destroying Roads time limit per test 2 seconds memory limit per test 256 megabytes inputstandard input outputstandard output 问题描述 In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbe…

题目:有n个城镇,m条边权为1的双向边让你破坏最多的道路,使得从s1到t1,从s2到t2的距离分别不超过d1和d2. #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <stack> #in…

Codeforces Round #531 (Div. 3) 题目总链接:https://codeforces.com/contest/1102 A. Integer Sequence Dividing 题意: 给一个数n,然后要求你把1,2.....n分为两个集合,使得两个集合里面元素的和的差的绝对值最小. 题解: 分析可以发现,当n%4==0 或者 n%3==0,答案为0:其余答案为1.之后输出一下就好了. 代码如下: #include <bits/stdc++.h> using name…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

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…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…