A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you…

A. Appleman and Easy Task time limit per test  1 second memory limit per test  256 megabytes input  standard input output  standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can…

A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you…

An Easy Task Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Ignatius was born in a lea…

An Easy Task Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4088 Accepted Submission(s): 2327   Problem Description Ignatius was born in a leap year, so he want to know when he could hold his bir…

E - Easy Task Description Calculating the derivation of a polynomial is an easy task. Given a function f(x) , we use (f(x))' to denote its derivation. We use x^n to denote xn. To calculate the derivation of a polynomial, you should know 3 rules:(1) (…

B. An Easy Task Time Limit: 1000ms Case Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer IO format: %lld      Java class name: Main Submit Status PID: 35999 Font Size:  +   - You are given an easy task by your supervisor -- to find the best va…

An Easy Task Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17062    Accepted Submission(s): 10902 Problem Description Ignatius was born in a leap year, so he want to know when he could hold h…

Problem Description Ignatius was born in a leap year, so he want to know when he could hold his birthday party. Can you tell him? Given a positive integers Y which indicate the start year, and a positive integer N, your task is to tell the Nth leap y…

Problem Description Ignatius was born in a leap year, so he want to know when he could hold his birthday party. Can you tell him? Given a positive integers Y which indicate the start year, and a positive integer N, your task is to tell the Nth leap y…

题意 每次把序列中最大的数a的一个和最小的数b的一个变成a-b.求最后是否能使序列里的数全部相同,能则输出这个相同的数. 分析 一定是有解的,不断减少最大数的个数,最大数减少为0个时,就是减少了不同数字的个数,最后就只剩一种数字了. 可以直接暴力模拟. 代码 #include<cstdio> int ok(int a[],int n) { int i; for(i=1; i<n; i++) if(a[i]!=a[i+1])break; return i==n; } int main()…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2969 全场第一水题.我不知道怎么解释,看代码就好了... #include<bits/stdc++.h> using namespace std; int main() { ]; cin>>t; while(t--) { cin>>n; ) { cin>>a[]; cout<<n<<endl; } else {…

水题,如题. #include <stdio.h> #define chk(Y) (Y%4==0 && Y%100!=0) || Y%400==0 int main() { int case_n; int i, y, n; scanf("%d", &case_n); while (case_n--) { scanf("%d %d", &y, &n); i = ; ) { if ( chk(y) ) i++; if…

题解:枚举即可…… #include <cstdio> int main(){ int now,y,n,T,count; scanf("%d",&T); while(T--){ scanf("%d%d",&y,&n); for(now=y;;now++){ if((now%4==0&&now%100!=0)||(now%400)==0)n--; if(!n){printf("%d\n",now)…

原文链接https://www.cnblogs.com/zhouzhendong/p/CF461D.html 题解 首先我们可以发现如果确定了第一行,那么方案就唯一了. 然后,我们来看看一个点的值确定了会导致什么: 假设我们确定了红色点的值,那么所有包含橙色的格子xor起来就等于红色格子的值,绿色蓝色也是. 第一排就比较特殊了. 如果我们在对第一排奇偶分类之后,如果我们可以得到第一行的前缀xor之间的关系,那么我们就有希望解决这个问题. 再看一种情况: 类似地意思,我们会发现这个东西遇到墙会反射…

http://acm.hdu.edu.cn/game/entry/problem/list.php?chapterid=1§ionid=2 1.2.5 #include<stdio.h> /* 题意:找闰年. if((i%4==0 && i%100!=0) || i%400==0)count++; 3 2005 25 1855 12 2004 10000 2108 1904 43236 */ int main() { int t,y,n; int i,count=0; whil…

假设该矩形是aij,那么有a(i,j)=a(i-1,j-1)^a(i-1,j+1)^a(i-2,j),不断递归下去可以发现a(i,j)=a(1,y-x+1)^a(1,y-x+3)^--^a(1,x+y-1). 那么,对第一行处理前缀和,Si=S(i-2)^a(1,i),即给出了两个数S的异或,只需将每一个点裂点为i和i',然后若Si^Sj=0,并查集上连边(i,j)(i',j'),否则连(i,j')(i',j),最后只需判断i和i'是否相连,相连即为0(这个可以理解为i表示i上是1,i'表示i上…

题目链接 A. Appleman and Easy Task time limit per test:2 secondsmemory limit per test:256 megabytesinput:standard inputoutput:standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can…

昨天没打,今天写了一下,前三题都没有难度吧. A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know h…

http://codeforces.com/contest/462 A:Appleman and Easy Task 要求是否全部的字符都挨着偶数个'o' #include <cstdio> using namespace std; ][]; int n; ][]; ]={,,-,}; ]={,-,,}; int main(){ scanf("%d",&n); gets(maz[]); ;i<n;i++){ gets(maz[i]); } ;i<n;i+…

比赛链接:http://codeforces.com/contest/462 这次比赛的时候,刚刚注冊的时候非常想好好的做一下,可是网上喝了个小酒之后.也就迷迷糊糊地看了题目,做了几题.一觉醒来发现rating掉了非常多,那个心痛啊! 只是.后来认真的读了题目,发现这次的div2并非非常难! 官方题解:http://codeforces.com/blog/entry/13568 A. Appleman and Easy Task 解析:         一个水题,推断每一个细胞周围是否都是有偶数…

称号: A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can…

Problem D"Decoding Task" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 Description In the near future any research and publications about cryptography are outlawed throughout the world on the grounds of national se…

  Code understanding is a task we are always doing, though we are not even aware that we're doing it. It's an obvious need when a colleague is leaving and another gets his code. However, we have to learn code in lots of cases, i.e. when: we continue…

Problem description Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it: d…

Instant Complexity Time Limit: 1000MS Memory Limit: 10000K Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm that runs in linear time is usually much f…

现在网络上又有越来越多的免费的(JS 图表)JavaScript图表图形绘制插件.我之前给一家网站做过复杂的图形,我们用的是 highchart.在那段时间,没有很多可供选择的插件.但现在不同了,很容易就可以找到很多功能非常不错的图表库.个人而言,此类插件的上升,是因为: 1. Flash 过去是最佳解决方案,但很多人多在从那迁移: 2. 现代浏览器及其更强大的计算能力,使其在转化绘制实时数据方面的能力更佳: 3. 绘制矢量图的不同技术愈发成熟:VML.SVG 和 Canvas. 当然了,你需要…

1. Introduction The Saga of Ryzom is a persistent massively-multiplayer online game (MMORPG) released in September 2004 throughout Europe and North America, localised in 3 languages so far. It has been developed by Nevrax since 2000, and was taken ov…

Best HTML5 frameworks are most popular because with the use of these frameworks you can create website with less effort and time. HTML5 language has very useful and amazing Features, but a few parts are not boosted by this language, so we required HT…

Cross platform support and flexible runtime engine(跨平台支持和灵活的运行时引擎) ASP.NET Core 1.0 offers support for cross-platform and this is seen to be an exciting new feature. It's easier for you to run your ASP.NET Application on Mac, Linux and Windows operat…