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…

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) (…

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…

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…

http://poj.org/problem?id=2826 An Easy Problem?! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10505   Accepted: 1584 Description It's raining outside. Farmer Johnson's bull Ben wants some rain to water his flowers. Ben nails two woode…

转载:http://www.klogk.com/posts/hdu4565/ 这里写的非常好,看看就知道了啊. 题意很easy.a,b,n都是正整数.求 Sn=⌈(a+b√)n⌉%m,(a−1)2<b<a2 这个题目也是2008年Google Codejam Round 1A的C题. 做法事实上很easy.记(a+b√)n为An,配项 Cn=An+Bn=(a+b√)n+(a−b√)n 两项恰好共轭,所以Cn是整数. 又依据限制条件 (a−1)2<b<a2⇒0<a−b√<…

Description 为了使得大家高兴,小Q特意出个自认为的简单题(easy)来满足大家,这道简单题是描述如下:有一个数列A已知对于所有的A[i]都是1~n的自然数,并且知道对于一些A[i]不能取哪些值,我们定义一个数列的积为该数列所有元素的乘积,要求你求出所有可能的数列的积的和 mod 1000000007的值,是不是很简单呢?呵呵! Input 第一行三个整数n,m,k分别表示数列元素的取值范围,数列元素个数,以及已知的限制条数.接下来k行,每行两个正整数x,y表示A[x]的值不能是y.…

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…

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()…

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)…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4864 题目大意: 有n个机器,m个任务.每个机器至多能完成一个任务.对于每个机器,有一个最大运行时间xi和等级yi,对于每个任务,也有一个运行时间xj和等级yj.只有当xi>=xj且yi>=yj的时候,机器i才能完成任务j,并获得500*xj+2*yj金钱.问最多能完成几个任务,当出现多种情况时,输出获得金钱最多的情况. 思路: 由于x的权重远大于y的权重,因此我们将工作按照x降序排列, x相同…

leetcode 34 最早出现和最后出现 class Solution { public int[] searchRange(int[] nums, int target) { int []ans={-1,-1} ; for(int i=0;i<nums.length;i++){ if(nums[i]==target){ ans[0]=i; break; } } for(int j=nums.length-1;j>=0;j--){ if(nums[j]==target){ ans[1]=j;…

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…

[python3]参数中的冒号与箭头 冒号后面是建议传入的参数类型 箭头后面是建议函数返回的类型…

Instant Complexity Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2017   Accepted: 698 Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm…

题目传送: Hybrid Crystals Problem Description > Kyber crystals, also called the living crystal or simply the kyber, and known as kaiburr crystals in ancient times, were rare, Force-attuned crystals that grew in nature and were found on scattered planets…

提高自己的实力, 也为了证明, 开始板刷lightoj,每天题量>=1: 题目的类型会在这边说明,具体见分页博客: SUM=54; 1000 Greetings from LightOJ [简单A+B] 1001 Opposite Task  [简单题] 1002 Country Roads[搜索题] 1003 Drunk[判环] 1004 Monkey Banana Problem [基础DP] 1006 Hex-a-bonacci[记忆化搜索] 1008 Fibsieve`s Fantabu…

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…

按照Leetcode的Tag来刷题,从easy到hard刷题 关于如何让Leetcode按难易程度排序,可按以下步骤: 1. 进入Leetcode后,点击code 2.点击code后,可查看所有题目,可看到右下角有Tag标志,选择想要刷的Tag即可 3. 点击Tag后,可看到该Tag下所有的题目,点击Acceptance, 可看到题目按从接收程度排序,然后点击两次Difficulty,则每次都会显示相同的排序方式 这样就可以尽情的从易到难刷题了,附上我的Leetcode刷题记录,求轻喷 Leet…

Cyclic Nacklace Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14416    Accepted Submission(s): 6016 Problem Description CC always becomes very depressed at the end of this month, he has checke…

Bzoj 3450: Tyvj1952 Easy 这里放上题面,毕竟是个权限题(洛谷貌似有题,忘记叫什么了) Time Limit: 10 Sec Memory Limit: 128 MB Submit: 1031 Solved: 772 Description 某一天WJMZBMR在打osu~~~但是他太弱逼了,有些地方完全靠运气:( 我们来简化一下这个游戏的规则 有n次点击要做,成功了就是o,失败了就是x,分数是按comb计算的,连续a个comb就有aa分,comb就是极大的连续o. 比如o…

Hdu 4864(Task 贪心) 原题链接 题意:给定n台机器和m个任务,任务和机器都有工作时间值和工作等级值,一个机器只能执行一个任务,且执行任务的条件位机器的两个值都大于等于任务的值,每完成一个任务可获奖励500x+2y.求完成最多任务前提下可获得最高奖励,输出任务数和奖励数. 解法:对机器和任务进行排序,x和y都从大到小排.循环考察每一个任务,把满足每一个任务x值的机器等级都记录下来,然后用满足任务等级的最小等级机器完成此任务.这样的做法可以保证: 前面的任务执行比后面的任务可获的奖励更…