F - Laurenty and Shop Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 586B Description A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hu…

B. Laurenty and Shop time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants t…

B. Laurenty and Shop Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/problem/B Description A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sa…

                                                                             B. Laurenty and Shop   A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sausage and cheese sand…

B. Laurenty and Shoptime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputA little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sa…

题目链接:http://codeforces.com/contest/586/problem/B B. Laurenty and Shop time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A little boy Laurenty has been playing his favourite game Nota for quit…

原题链接:http://codeforces.com/contest/586/problem/B 题意: 大概就是给你一个两行的路,让你寻找一个来回的最短路,并且不能走重复的路. 题解: 就枚举上下选的是哪条路来跨过主干道就好 代码: #include<iostream> #include<cstring> #include<vector> #define MAX_N 55 using namespace std; int n; ][MAX_N]; ][MAX_N];…

E. Ladies' Shop time limit per test 8 seconds memory limit per test 256 megabytes input standard input output standard output A ladies' shop has recently opened in the city of Ultima Thule. To get ready for the opening, the shop bought n bags. Each b…

题意:给定 n 个工作的最好开始时间,和持续时间,现在有两种方法,第一种,如果当前的工作能够恰好在最好时间开始,那么就开始,第二种,如果不能,那么就从前找最小的时间点,来完成. 析:直接暴力,每次都先去看看能不能在最好时间完成,如果不能,就去找最小的时间点. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #incl…

A. Alena's Schedule 间隔0长度为1被记录  1被记录  其余不记录 #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <vector> using namespace std; int main() { int t; cin>>t; ; ,ans= ; bool mk = false; whi…