Magic Pen 6
At the end of this term, teacher gives Timer a job to deliver the list of N students who fail the course to dean's office. Most of these students are Timer's friends, and Timer doesn't want to see them fail the course. So, Timer decides to use his magic pen to scratch out consecutive names as much as possible. However, teacher has already calculated the sum of all students' scores module M. Then in order not to let the teacher find anything strange, Timer should keep the sum of the rest of students' scores module M the same.
Plans can never keep pace with changes, Timer is too busy to do this job. Therefore, he turns to you. He needs you to program to "save" these students as much as possible.
The first line of each case contains two integer N and M, (0< N <= 100000, 0 < M < 10000),then followed by a line consists of N integers a1,a2,...an (-100000000 <= a1,a2,...an <= 100000000) denoting the score of each student.(Strange score? Yes, in great HIT, everything is possible)
1 6
3 3
2 3 6
2 5
1 3
2
0
The magic pen can be used only once to scratch out consecutive students.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm> using namespace std; const int N=1e5+; long long num[N],sum[N]; int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
sum[]=;
for(int i=;i<=n;i++){
scanf("%lld",&num[i]);
sum[i]=sum[i-]+num[i]%m;
}
int ans=,flag=;
for( int i=n; i>&&flag; i-- ){
for( int j=n; j>=i; j-- ){
if((sum[j]-sum[j-i])%m==){
ans=i;
flag=;
break;
}
}
}
cout<<ans<<endl;
}
return ;
}
Magic Pen 6的更多相关文章
- HDU 4648 Magic Pen 6 (。。。。。。。。。。)
Magic Pen 6 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total ...
- hdu 4648 - Magic Pen 6(“水”题)
摘自题解: 题意转化一下就是: 给出一列数a[1]...a[n],求长度最长的一段连续的数,使得这些数的和能被M整除. 分析: 设这列数前i项和为s[i], 则一段连续的数的和 a[i]+a[i+1] ...
- HDU 4648 Magic Pen 6
题目链接 6Y什么水平.. #include <cstdio> #include <cstring> #include <string> #include < ...
- HDU 4648 Magic Pen 6 思路
官方题解: 题意转化一下就是: 给出一列数a[1]...a[n],求长度最长的一段连续的数,使得这些数的和能被M整除. 分析: 设这列数前i项和为s[i], 则一段连续的数的和 a[i]+a[i+1] ...
- HDU-4648 Magic Pen 6 简单题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4648 求遍前缀和,然后扫描标记下就可以了... //STATUS:C++_AC_453MS_1792K ...
- 【 2013 Multi-University Training Contest 5 】
HDU 4647 Another Graph Game 如果没有边的作用,显然轮流拿当前的最大值即可. 加上边的作用,将边权平均分给两个点,如果一个人选走一条边的两个点,就获得了边的权值:如果分别被两 ...
- 《割绳子》《蜡笔物理学》《Contre Jour》《顽皮鳄鱼爱洗澡》等游戏用Box2D引擎实现物理部分的方法(转)
从最热门游戏排行榜和flash游戏网站上,你能看到什么?许多2D游戏都有非常出色的物理学和美术设计.现在我们要学习那些游戏使用了什么物理学以及如何用Box2D制作它们. 除了知道是“什么”,更重要的是 ...
- Codeforces CF#628 Education 8 D. Magic Numbers
D. Magic Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
- [8.3] Magic Index
A magic index in an array A[0...n-1] is defined to be an index such that A[i] = i. Given a sorted ar ...
随机推荐
- Groovy 设计模式 -- 责任链模式
Chain of Responsibility Pattern http://groovy-lang.org/design-patterns.html#_chain_of_responsibility ...
- 爬取qq音乐巅峰榜---内地音乐的榜单
import requestsimport jsonimport sys for i in range(0,10): url = "https://szc.y.qq.com/v8/fcg-b ...
- 【Unity]】AR小工具-Vuforia
很有意思的增强现实玩具,六分钟应用. https://www.youtube.com/watch?v=khavGQ7Dy3c
- Coursera, Big Data 2, Modeling and Management Systems (week 1/2/3)
Introduction to data management 整个coures 2 是讲data management and storage 的,主要内容就是分布式文件系统,HDFS, Redis ...
- BIO 和 NIO
一.阻塞(Block)和非阻塞(NonBlock) 阻塞和非阻塞是进程在访问数据的时候,数据是否准备就绪的一种处理方式,当数据没有准备的时候阻塞: 阻塞:往往需要等待缞冲区中的数据准备好过后才处理其他 ...
- 帆软报表(finereport) 动态报表
动态表实现了不同的人根据需要选择不同的表进行查看,从而提高查询效率 在定义数据集时,通过if函数来判断参数的值从而来实现调用不同的数据表 如直接将SQL语句定义成:SELECT * FROM ${if ...
- 【原创】大数据基础之Azkaban(1)简介、源代码解析
Azkaban3.45 一 简介 1 官网 https://azkaban.github.io/ Azkaban was implemented at LinkedIn to solve the pr ...
- VUE 多页面配置(二)
1. 概述 1.1 说明 项目开发过程中会遇到需要多个主页展示情况,故在vue单页面的基础上进行配置多页面开发以满足此需求,此记录为统一配置出入口. 2. 实例 2.1 页面配置 使用vue脚手架搭建 ...
- 六.ansible批量管理服务
期中集群架构-第六章-ansible批量管理服务介绍====================================================================== 01. ...
- python3 字典(dictionary)(一)
一.定义:是另一种可变容器模型,可存储任意类型对象:(也被称为关联数组或哈希表:存储的数据是没有顺序的) 语法为: d = {key1 : value1, key2 : value2 } #----- ...