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Fox and Cross Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a s…
Rebranding Problem Description The name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding - an active marketing strategy, that includes a set of measures to change either the bra…
B. Cards time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can…
C. Report time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodi…
B. Barnicle time limit per test: 1 second memory limit per test :256 megabytes input: standard input output: standard output Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know th…
活生生打成了大模拟... #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int>PII; const double eps=1e-5; const double pi=acos(-1.0); //const int mod=1e9+7; const int INF=0x3f3f3f3f; //ht…
题目链接:http://codeforces.com/problemset/problem/719/C 题目大意: 留坑...…
题意:给定三个操作,1,是x应用产生一个通知,2,是把所有x的通知读完,3,是把前x个通知读完,问你每次操作后未读的通知. 析:这个题数据有点大,但可以用STL中的队列和set来模拟这个过程用q来标记是哪个应用产生的,用set来记录是第几个通知. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include &l…
强行模拟 纪念一下…… #include<stdio.h> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<map> #include<vector> #include<queue> #define M(a,b) memset(a,b,sizeof…
题目大意:给出n个点的坐标,和你当前的坐标,求走过n-1个点的最短路程. 题目思路:走过n-1个点,为了使路程更短,那么不走的点只可能第一个点或最后一个点.模拟就行了,比较恶心. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h> #include<stdlib.h> #include<queue&…