B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

The Meeting Place Cannot Be Changed Problem Description The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the…

The Meeting Place Cannot Be Changed 我发现我最近越来越zz了,md 连调程序都不会了,首先要有想法,之后输出如果和期望的不一样就从输入开始一步一步地调啊,tmd现在输出不一样就知道在那等着,有毛用啊. [题目链接]The Meeting Place Cannot Be Changed [题目类型]二分答案 &题解: 二分时间,判断函数比较难想,要先假设所有人都向左走,找到坐标最大的那个,之后在假设所有人都向右走,找到坐标最小的那个,如果最大的<=最小的就行…

地址:http://codeforces.com/contest/782/problem/B 题目: B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from…

780B - The Meeting Place Cannot Be Changed 思路: 二分答案: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-7 #define INF 1e18 #define maxn 60005 int n; double xi[maxn]…

题目链接 The Meeting Place Cannot Be Changed 二分答案即可. check的时候先算出每个点可到达的范围的区间,然后求并集.判断一下是否满足l <= r就好了. eps我设了1e-7. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for(int i(a); i <= (b); ++i) + ; ; double mi, ma; struct node{ dou…

                                                               B. The Meeting Place Cannot Be Changed The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost build…

https://vjudge.net/problem/CodeForces-782B B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to…

B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

1.debug,全部打印 2.打断点debug,出现单步调试等按钮,只运行断点前 3.setup over 调试一行代码 4.setup out 运行断点后面所有代码 5.debug窗口显示调试按钮 6.运行到对应的点会显示变量的值 7.step into:单步执行,遇到子函数就进入并且继续单步执行(简而言之,进入子函数): step over:在单步执行时,在函数内遇到子函数时不会进入子函数内单步执行,而是将子函数整个执行完再停止,也就是把子函数整个作为一步.有一点,经过我们简单的调试,在不存…

题目链接 大致题意 把一个图分成三块,要求任意两块之间是完全图,块内部没有连线 分析 首先根据块内没有连线可以直接分成两块 假定点1是属于块1的,那么所有与点1连接的点,都不属于块1:反之则是块1的 然后在所有不属于块1的点内随意找一点k,设定其属于块2,那么所有与点k连接的点且不属于块1,则是块3. 块分完了,然后是判断每个块是否满足条件,我通过下面三条来判断 1.每个块都有点 2.每个块内部没有连线,即没有一条线的两个端点在同一个块内 3.每个块内的点的度等于其他两个块的点个数和也等于n减去…

There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebo…

C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…

B. Game of Credit Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and deci…

B. Intersection time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined…

Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and mcolumns. Let number a[i][j] represents the calories burned by performing workout at the…

算是一题普通数论+思维题吧. 大概很多人是被题意绕晕了. 思路: 首先常规操作求出X的质因子. 然后题目要求的是,X的每个质因子p,在g(i,p)的连乘.i∈[1,n]: 我们转换下思维,不求每一个g(i,p)中最终是哪些 p的幂次,而是反求 每个p的幂次对结果的贡献. 显而易见,p^k在1~n的出现的次数就是  [n/(p^k)]. 这样枚举所有质因子,计算中再利用快速幂取模便可以得到答案 //#pragma comment(linker, "/STACK:1024000000,1024000…

time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmo…

Link 题意:给出$n$个坐标$x_i$,$n$个速度$v_i$问使他们相遇的最短时间是多少. 思路:首先可肯定最终相遇位置必定在区间$[0,max(x_i)]$中,二分最终位置,判断左右部分各自所花时间,缩小范围即可. /** @Date : 2017-05-09 22:07:43 * @FileName: 782B.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@gmail.com) * @Link : https://gi…

三分显然,要注意EPS必须设成1e-6,设得再小一点都会TLE……坑炸了 #include<cstdio> #include<algorithm> #include<cmath> using namespace std; #define EPS 0.000001 int n,x[60010],v[60010]; double calc(double p) { double res=0; for(int i=1;i<=n;++i) res=max(res,fabs(…

题意:题意:给出n个人的在x轴的位置和最大速度,求n个人相遇的最短时间. 析:二分时间,然后求并集,注意精度,不然会超时. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream>…

题意: The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the road there are n friends, and i-th of them is standi…

链接:http://codeforces.com/problemset/problem/782/B 题意: N个点,需要找到一个点使得每个点到这个点耗时最小,每个点都同时开始,且都拥有自己的速度 题解: 对于一个确定的位置,如果耗时最久的点在右边,则这个位置可以往右靠,否则就往左靠,这样,一个二分的解法就形成了 import java.lang.Math; import java.util.Scanner; public class CodeForces_403_B { private stat…

[题目链接]:http://codeforces.com/contest/782/problem/B [题意] 每个人都有一个速度,只能往上走或往下走; 然后让你找一个地方,所有人都能够在t时间内到达; 让t最小. [题解] 很明显的二分了; 二分时间t; 对于某个时间t,这个人都有一个能够到达的唯一区间 [x[i]-v*t..x[i]+v*t] 如果所有人的区间都有交集. 那么就能在那些交集里面随便选一个点了; 区间的交集可以用 最大的左端点小于等于最小的右端点这个依据来判断是否存在 想让答案…

http://codeforces.com/contest/782/problem/B 题意:有n个人,每个人有一个位置和速度,现在要让这n个人都走到同一个位置,问最少需要的时间是多少. 思路:看上去很像二分搜索啊!枚举距离,判断是否有更少的时间,然后发现时间不随着距离单调增减,想起前两天被三分虐了一道题,那就是三分吧. 三分枚举距离,然后判断是否有更少的时间就可以了.比赛时候用了max函数还有精度开太大,超时了,索性直接循环100次. #include <bits/stdc++.h> usi…

题意:给你一张有向图,某人会任意选择起点然后走无穷多步,问是否存在一个点(要求输出)不管他起点在何处怎么走都必经?n<=100005,m<=500005. 标程: #include<bits/stdc++.h> using namespace std; ; int cnt,vis[N],n,m,u,v,a[N],tried[N],head[N]; ]; void add(int x,int y) {num[++cnt].to=y;num[cnt].next=head[x];head…

看来快掉到灰名的蒟蒻涨rating也快... A题模拟一下就好(一开始还sb,, #include<bits/stdc++.h> #define LL long long using namespace std; ]; ],n; int main() { scanf("%d",&n); ; i<=*n; i++) { int x; scanf("%d",&x); ; else tot--; ans=max(ans,tot); } c…

题意:在一维坐标轴上,给定n个点的坐标以及他们的最大移动速度,问他们能聚到某一点处的最短时间. 分析: 1.二分枚举最短时间即可. 2.通过检查当前时间下,各点的最大移动范围之间是否有交集,不断缩小搜索范围. 3.相当于二分枚举找右临界线,符合要求的点都在右边. 4.通过给二分一个查找次数的上界,eg:k=200,而不是dcmp(l, r)<=0,后者会tle. 5.检查是否有交集,就是以第一个点的最大移动区间为标准,不断与其他区间取交集并更新再继续比较. #include<cstdio>…

1.利用 .htaccess 防止盗链 如果不喜欢别人在他们的网页上链接自己的图片.文档的话,也可以通过htaccess的指令来做到.当然这样也可以对你的网站服务器压力变小! 这次先给出‘代码’,然后进行详细的讲解!这个东西纠结了我很久啊,既然自己懂一些,就拿出来和大家一起分享一下,可能有些地方理解错误了,希望发现不对的‘童鞋’能指出,共同进步! RewriteCond %{HTTP_REFERER} !^$ RewriteCond %{HTTP_REFERER} !^http://localh…

加载Rewrite模块: 在conf目录下httpd.conf中找到 LoadModule rewrite_module modules/mod_rewrite.so 这句,去掉前边的注释符号“#”,或添加这句. 允许在任何目录中使用“.htaccess”文件,将“AllowOverride”改成“All”(默认为“None”):  代码如下 复制代码 # AllowOverride controls what directives may be placed in .htaccess file…